
Class TK^lU 
Book. .?> 5 •'- 



Copyright^ . 



\ 9Q % 



COPYRIGHT DEPOSIT; 



The D. Van Nostrand- Company 

intend this book to be sold to the Public 
at the advertised price, and supply it to 
the Trade on terms which will not allow 
of discount. 



Alternating-Current Machines 

BEING THE SECOND VOLUME OF 

DYNAMO ELECTRIC MACHINERY 

ITS CONSTRUCTION, DESIGN, 
AND OPERATION 



BY 

SAMUEL SHELDON, A.M., Ph.D., D.Sc. 

PROFESSOR OF PHYSICS AND ELECTRICAL ENGINEERING AT THE POLYTECHNIC 

INSTITUTE OF BROOKLYN AND PAST-PRESIDENT OF THE AMERICAN 

INSTITUTE OF ELECTRICAL ENGINEERS 

AND 

HOBART MASON, B.S., E.E. 

ASSOCIATE OF THE AMERICAN INSTITUTE OF ELECTRICAL ENGINEERS 
AND 

ERICH HAUSMANN, B.S. E.E. 

INSTRUCTOR IN ELECTRICAL ENGINEERING AT THE POLYTECHNIC 
INSTITUTE OF BROOKLYN, AND ASSOCIATE OF THE AMERICAN 
INSTITUTE OF ELECTRICAL ENGINEERS 

SEVENTH EDITION 

COMPLETEL V REWRITTEN 



NEW YORK 

D. VAN NOSTRAND COMPANY 

23 Murray and 27 Warren Sts. 

LONDON 

CROSBY LOCKWOOD & SON 

7 Stationers' Hall Court, Ludgate Hill 

1908 



fuBRARYofCONfiRESSi 
&wo tiooies rtecewtfc 

OCT 3 WW 






Copyright, 1908, by 
D. VAN NOSTRAND COMPANY 



Stanbope lPtess 

F. H 9ILSON COMPANY 
BOSTON. U.S.A. 



PREFACE TO FIRST EDITION, 



This book, like its companion volume on Direct Current 
Machines, is primarily intended a's a text-book for use in 
technical educational institutions. It is hoped and be- 
lieved that it will also be of use to those electrical, civil, 
mechanical, and hydraulic engineers who are not perfectly 
familiar with the subject of Alternating Currents, but whose 
work leads them into this field. It is furthermore intended 
for use by those who are earnestly studying the subject 
by themselves, and who have previously acquired some 
proficiency in mathematics. 

There are several methods of treatment of alternating- 
current problems. Any point is susceptible of demonstra- 
tion by each of the methods. The use of all methods in 
connection with every point leads to complexity, and is 
undesirable in a book of this character. In each case that 
method has been chosen which was deemed clearest and 
most concise. No use has been made of the method of 
complex imaginary numbers. 

A thorough understanding of what takes place in an 
alternating-current circuit is not to be easily acquired. It 
is believed, however, that one who has mastered the first 
four chapters of this book will be able to solve any practi- 
cal problem concerning the relations which exist between 
power, electro-motive forces, currents, and their phases in 



IV PREFACE. 

series or multiple alternating-current circuits containing 
resistance, capacity, and inductance. 

The next four chapters are devoted to the construction, 
principle of operation, and behavior of the various types of 
alternating-current machines. Only American machines 
have been considered. 

A large amount of alternating-current apparatus is used 
in connection with plants for the long-distance transmission 
of power. This subject is treated in the ninth chapter. 
The last chapter gives directions for making a variety of 
tests on alternating-current circuits and apparatus. 

No apology is necessary for the introduction of cuts and 
material supplied by the various manufacturing companies. 
The information and ability of their engineers, and the taste 
and skill of their artists, are unsurpassed, and the informa- 
tion supplied by them is not available from other sources. 
For their courteous favors thanks is hereby given. 



PREFACE TO THE SEVENTH EDITION. 



Tb:e extensive adoption of this volume as a text-book 
for the use of students on other than electrical courses and 
the growing tendency, in many Institutions, to require 
more thorough and extended work in electrical subjects 
from such students, have determined the scope of the 
present revision. In those cases where insufficient time is 
available for covering all the ground contained herein, it 
will be found that portions, which the instructor will 
probably desire to omit, are so treated that the remainder 
will constitute a coordinated treatment. It is also believed 
that, in the majority of Institutions, the book as a whole 
will be found adapted for the use of students on electrical 
courses. The manner of presentation is in many parts 
different from that which would be employed in a book 
written for engineers, but an extended experience in 
teaching young men of average attainments has proved it 
to be effective. As a student seldom gets a thorough 
understanding of a subject of this character without 
making numerical computations, problems have been 
introduced at the conclusion of each chapter. 



CONTENTS 



CHAPTER I. 



Properties of Alternating Currents. 

ART PAGE 

i. Definition of an Alternating Current -i 

2. Frequency i 

3. Wave-Shape 3 

4. Distortion 5 

5. Effective Values of E.M.F. and of Current 7 

6. Form Factor of Non-Sine Curves 10 

7. Phase 12 

8. Power in Alternating-Current Circuits 14 

9. Non-Sine Waves 18 

10. E.M.F.'s in Series 20 

Problems 25 



CHAPTER II. 

Self-Induction. 

11. Self-Inductance 26 

12. Unit of Self-Inductance 27 

13. Practical Values of Inductances 29 

14. Things which influence the Magnitude of L 30 

15. Formula? for calculating Inductances 31 

16= Growth of Current in an Inductive Circuit 33 

17. Decay of Current in an Inductive Circuit 34 

18. Magnetic Energy of a started Current 36 

19. Current produced by a Harmonic E M.F. in a Circuit having Resist- 

ance and Inductance 37 

20. Instantaneous Current produced by a Harmonic E.M.F. in a Circuit 

having Resistance and Inductance 41 

21. Choke Coils 44 

Problems ..:.'. 47 

vii 



viii CONTENTS. 



CHAPTER III, 

Capacity. 

ART. PAGE 

22. Condensers 48 

23. Capacity Formulae 53 

24. Connection of Condensers in Parallel and in Series 55 

25. Decay of Current in a Condensive Circuit 57 

26. Energy stored in Dielectric „ 60 

27. Condensers in Alternating-Current Circuits — Hydraulic Analogy. 60 

28. Phase Relations . . . „ 61 

29. Current and Voltage Relations 63 

30. Instantaneous Current in a Circuit having Capacity and Resistance 65 
Problems 68 



CHAPTER IV. 

Alternating-Current Circuits. 

31. Resistance, Inductance, and Capacity in an Alternating-Current 

Circuit 70 

32. Definitions of Terms 71 

^^. Representation of Impedance and Admittance by Complex 

Numbers 74 

34. Instantaneous Current in a Circuit having Inductance, Capacity and 

Resistance 76 

35. Resonance 80 

36. Damped Oscillations 82 

37. Polygon of Impedances 83 

38. A Numerical Example applying to the Arrangement shown in 

Fig- 50 • • • .'. 85 

39. Polygon of Admittances 87 

40. Impedances in Series and in Parallel 90 

Problems 92 

CHAPTER V. 

Alternators. 

41. Alternators 94 

42. Electromotive Force generated 96 

43. Armature Windings. . . . „ 99 

44. Voltage and Current Relations in Two-phase Systems . . . . „ .... . 101 

45. Voltage and Current Relations in Three-phase Systems 104 



CONTENTS. ix 

ART. PAGE 

46. Voltage and Current Relations in Four-Phase Systems 106 

47. Measurement of Power 107 

48. Saturation 113 

49. Regulation „ 116 

50. E.M.F. and M.M.F. Methods of calculating Regulation 119 

51. Regulation for Constant Potential 124 

52. Efficiency 133 

53. Rating 135 

54. inductor Alternators 136 

55. Revolving Field Alternators , 139 

56. Self- Exciting Alternators 145 

Problems . . 146 



CHAPTER VI. 
The Transformer. 

57. Definitions 149 

58. The Ideal Transformer 151 

59. Core Flux 154 

60. Transformer Losses 156 

61. Core Losses 156 

62. Exciting Current 159 

63. Equivalent Resistance and Reactance of a Transformer 163 

64. Copper Losses 165 

65. Efficiency 166 

66. Calculation of Equivalent Leakage Inductance 168 

67. Regulation 173 

68. Circle Diagram 179 

69. Methods of connecting Transformers 181 

70. Lighting Transformers 188 

71. Cooling of Transformers 192 

72. Constant Current Transformers 195 

73. Polyphase Transformers 198 

Problems 200 

CHAPTER VII. 

Motors. 

Induction Motors. 

74. Rotating Field 202 

75. The Induction Motor 203 

76. Starting of Squirrel-Cage Motors 207 



X CONTENTS. 

ART . PA G E 

77. Principle of Operation of the Induction Motor 210 

78. Relation between Speed and Efficiency 213 

79. Determination of Torque 214 

80. The Transformer Method of Treatment 215 

81. Leakage Reactance of Induction Motors 216 

82. Calculation of Exciting Current 227 

83. Circle Diagram by Calculation 231 

84. Circle Diagram by Test 233 

85. Performance Curves from Circle Diagram 236 

86. Method of Test with Load 238 

87. Phase Splitters 241 

88. The Single-Phase Induction Motor 242 

89. The Monocyclic System . '„- 244 

90. Frequency Changers 244 

91. Speed Regulation of Induction Motors 245 

92. The Induction Wattmeter 246 

Synchronous Motors. 

93. Synchronous Motors 249 

94. Special Case 252 

95. The Motor E.M.F 256 

96. Starting Synchronous Motors 258 

97. Parallel Running of Alternators 262 

Single-Phase Commutator Motors. 

98. Single- Phase Commutator Motors 262 

99. Plain Series Motor . 264 

100. Characteristics of Plain Series Motor 268 

101. Compensated Series Motors 270 

102. Sparking in Series Motors ...,-. 274 

103. Repulsion Motors „ 277 

104. Series-Repulsion Motor „ 280 

Problems . . e 282 

CHAPTER VIII. 

Converters. 

105. The Converter 284 

106. E.M.F. Relations 286 

107. Current Relations 288 

108. Heating of the Armature Coils 290 

109. Capacity of a Converter . . 291 



CONTENTS. XI 

ART. PAGE 

no. Starting a Converter 291 

in. Armature Reaction 291 

112. Regulation of Converters 293 

113. Mercury Vapor Converter 296 

Problems 299 

CHAPTER IX. 
Power Transmission. 

114. Superiority of Alternating Currents 300 

115. Frequency 302 

116. Number of Phases 304 

117. Voltage 305 

118. Economic Drop . . . . „ 306 

119. Line Resistance 309 

120. Line Inductance 310 

121. Line Capacity 313 

122. Regulation. 316 

123. Conductor Material 318 

124. Insulators 319 

125. Sag of Conductors 322 

126. Line Structure 326 

127. Spans and Layout 329 

128. Example of Design of Transmission Line 331 

Problem . 343 



ALTERNATING-CURRENT MACHINES. 



CHAPTER I. 

PROPERTIES OF ALTERNATING CURRENTS. 

i. Definition of an Alternating Current. — An alter- 
nating current of electricity is a current which changes 
its direction of flow at regularly recurring intervals. 
Between these intervals the value of the current may 
vary in any way. In usual practice, the value varies with 
some regularity from zero to a maximum, and decreases 
with the same regularity to zero, then to an equal max- 
imum in the other direction, and finally to zero again. In 
practice, too, the intervals of current flow are very short, 
ranging from ^ to ^-J-g- second. 

2. Frequency. — When, as stated above, a current has 
passed from zero to a maximum in one direction, to zero, 
to a maximum in the other direction, and finally to zero 
again, it is said to have completed one cycle. That is to 
say, it has returned to the condition in which it was first 
considered, both as to value and as to direction, and is 
prepared to repeat the process described, making a second 
cycle. It should be noted that it takes two alternations 
to make one cycle. The tilde ( ~ ) is frequently used to 
denote cycles. 



2 ALTERNATING-CURRENT MACHINES. 

The term frequency is applied to the number of cycles 
completed in a unit time, i.e., in one second. Occasionally 
the word alternations is used, in which case, unless other- 
wise specified, the number of alternations per minute is 
meant. Thus the same current is spoken of as having a 
frequency of 25, or as having 3000 alternations. The use 
of the word alternations is condemned by good practice. 
In algebraic notation the letter / usually stands for the 
frequency. 

The frequency of a commercial alternating current 
depends upon the work expected of it. For power a 
low frequency is desirable, particularly for converters. 
The great Niagara power plant uses a frequency of 25. 
Lamps, however, are operated satisfactorily only on fre- 
quencies of 50 or more. Early machines had higher 
frequencies, — 125 and 133 (16,000 alternations) being 
usual, — but these are almost entirely abandoned because 
of their increased losses and their unadaptability to the 
operation of motors and similar apparatus. 

In the Report of the Committee on Standardization of 
the American Institute of Electrical Engineers is the 
following: "In alternating-current circuits, the follow- 
ing frequencies are standard: 

25 ~ 

"These frequencies are already in extensive use, and 
it is deemed advisable to adhere to them as closely as 
possible." 

The frequency of an alternating current is always that 
of the E.M.F. producing it. To find the frequency of the 
pressure or the current produced by any alternating-cur- 



PROPERTIES OF ALTERNATING CURRENTS. 3 

rent generator, if V be the number of revolutions per 
minute, and / be the number of pairs of poles, then 

/ = / — 

3. Wave-shape If, in an alternating current, the 

instantaneous values of current be taken as ordinates, and 
time be the abscissae, a 
curve, as in Fig. 1, may be 
developed. The length of 
the abscissa for one com- 
1 




Fig. 



plete cycle is— seconds. 

Imagine a small cylinder, 
Fig. 2, carried on one end of a wire, and rotated uniformly 
about the other end in a vertical plane. Imagine a hori- 
zontal beam of parallel rays of light to be parallel to the 
plane of rotation, and to cast a shadow of the cylinder on 




a plane screen perpendicular to the rays. The shadow 
will move up and down, passing from the top of its travel 
to the bottom in a half revolution, and from the bottom 




4 ALTERNATING-CURRENT MACHINES. 

back to the top in another half revolution with a perfect 
harmonic motion. Now imagine the screen to be moved 
horizontally in its own plane with a uniform motion, and 
the positions of the shadow suitably recorded on it, — as 

on sensitized paper or on 
a photographic film, a 
slotted screen protecting 
all but the desired portion 
from exposure. Then the 
trace of the shadow will 
be as in Fig. 3. The 
abscissae of this curve 
may be taken as time, as in the preceding curve, the ab- 
scissa of one complete cycle being the time in seconds of 
one revolution. Or, with equal relevancy, the abscissae 
may be expressed in degrees. Consider the cylinder to be 
in a zero position when the radius to which it is attached 
is horizontal. Then the abscissa of any point is the angle 
which must be turned through in order that the cylinder 
may cast its shadow at that point. In this case the abscissa 
of a complete cycle will be 360 , or 2tt. Consideration of 
the manner in which the curve has been formed shows 
that the ordinate of any point is proportional to the sine 
of the abscissa of that point, expressed in degrees. Hence 
this is called a sinusoid or sine curve. 

If the maximum ordinate of this curve, which corresponds 
to the length of the moving radius, or OA in Fig. 4, repre- 
sents E m , then the instantaneous value of the voltage, E', 
at / seconds after the beginning of any cycle, will be AB, or 
E m sin 6. But, since OA traverses 2 n radians during one 
complete revolution, it will sweep over 2 nf radians per 
second, and, as angular velocity, represented by co } is the 




Fig. 4. 



PROPERTIES OF ALTERNATING CURRENTS. 5 

angle turned through in unit time, it follows that the angular 
velocity of OA is 2 nj. The angular velocity may also be 
expressed as d/t, or = cot 

= 2 Tljt. 

Hence E' = E m sin cot 

= E m sin 2 njt i 

which is equivalent to neglecting 
all those intervals of time cor- 
responding to whole cycles, and 
considering only the time elapsed since the end of the last 
completed cycle. In Fig. 4, OA is termed the radius 
vector, and 0, the vectorial angle or displacement. Graphic 
solutions of alternating-current problems may be effected 
by the use of vectors. 

4. Distortion. — The ideal pressure curve from an alter- 
nator is sinusoidal. Commercial alternators, however, do 
not generate true sinusoidal pressures. But the sine curve 
can be treated with relative simplicity, and the curves of 
practice approximate so closely to the sine form, that mathe- 
matical deductions based on sine curves can with propriety 
be applied to those of practice. Two of these actual curves 
are shown in Fig. 5. 

The shape of the pressure curve is affected by irregular 
distribution of the magnetic flux. Also uneven angular 
velocity of the generator will distort the wave-shape, 
making it, relative to the true curve, lower in the slow 
spots and higher in the fast ones. Again, the magnetic 
reluctance of the armature may vary in different angular 
positions, particularly if the inductors are laid in a few 
large slots. This would cause a periodic variation in the 



ALTERNATING-CURRENT MACHINES. 



reluctance of the whole magnetic circuit and a correspond- 
ing pulsation of the total magnetic flux. All these influ- 
ences operate at open circuit as well as under load. 



E.M.F. CURVE 

3 PHASE 

40 POLE 

2000 K.W. 

25 '^ 

FULLY LOADED 




There are two other causes which act to distort the 
wave-shape only when under load. For any separately 
excited generator, a change in the resistance or apparent 
resistance of the external circuit will cause a change in the 



PROPERTIES OF ALTERNATING CURRENTS. 7 

terminal voltage of the machine. As is explained later, 
the apparent resistance (impedance) of a circuit to alter- 
nating currents depends upon the permeability of the iron 
adjacent to the circuit. Permeability changes with mag- 
netization. Now, because an alternating current is flow- 
ing, the magnetization changes with the changing values 
of current. This, by varying the permeability, sets up a 
pulsation in the impedance and affects the terminal volt- 
age of the machine, periodically distorting the wave of 
pressure from the true sine. 

There are cases of synchronously pulsating resistances. 
The most common is that of the alternating arc. With 
the same arc the apparent resistance of the arc varies in- 
versely as the current. So when operated by alternating 
currents, the resistance of a circuit of arc lamps varies syn- 
chronously, and distorts the pressure wave-shape in a 
manner analogous to the above. 

Summing up, the wave-shape of pressure may be dis- 
torted : At open circuit as zvell as wider load ; by lack of 
uniformity of magnetic distribution, by pulsating of mag- 
netic field, by variation in angular velocity of armature ; 
and under load only ; by pulsation of impedance, by pulsa- 
tion of resistance. And the effects of any or all may be 
superimposed. 

5. Effective Values of E.M.F. and of Current One 

ampere of alternating current is a current of such instan- 
taneous values as to have the same heating effect in a con- 
ductor as one ampere of direct current. This somewhat 
arbitrary definition probably arose from the fact that alter- 
nating currents were first commercially employed in light- 
ing circuits, where their utility was measured by the heat 



8 



ALTERNATING-CURRENT MACHINES. 



they produced in the filaments ; and further from the fact 
that the only means then at hand of measuring alternating 
currents were the hot-wire instruments and the electro- 
dynamometer, either of which gives the same indication 
for an ampere of direct current or for what is now called 
an ampere of alternating current. 

The heat produced in a conductor carrying a current is 
proportional to the square of the current. In an alternat- 
ing current, whose instantaneous current values vary, the 
instantaneous rate of heating is not proportional to the 
instantaneous value, nor yet to the square of the average 

of the current values, but to the 
square of the instantaneous cur- 
rent value. And so the average 
heating effect is proportional to 
the mean of the squares of the 
instantaneous currents. 

The average current of a sinu- 
soidal wave of alternating current, whose maximum value 
is I m , is equal to the area of one lobe of the curve, Fig. 6, 




Fig. 6. 



divided by its base line ir. Thus 

x 



I m sin 6d0 






^[-cosC = -4 



But the heating value of such a current varies, as 



/ 2 = 



P 



I* sin 2 Odd 



I 2 

7T 



value of the current, I = 



V2 



sin 2 =-f m - 

|_2 4 Jo 2 

ntity is called the erf 
This has the same heating 



The square root of this quantity is called the effective 
L 



PROPERTIES OF ALTERNATING CURRENTS. 9 

effect as a direct current /, and the effective values are 
always referred to unless expressly stated otherwise. 
Alternating-current ammeters are designed to read in 
effective amperes. 

Since current is dependent upon the pressure, the 
resistance or apparent resistance of a circuit remain- 
ing constant, it is obvious that if / = -^ then ' does 

E 2 V2 

also E = ~- Likewise if average I = - I m then does also 

V2 «" 

2 
average E = - E m . Or these may be demonstrated in a 

TV 

manner analogous to the above. 

The maximum value of pressure is frequently referred 
to in designing alternator armatures, and in calculating 
dielectric strength of insulation. There have arisen vari- 
ous ways of indicating that effective values are meant, 
for instance, the expressions, sq. root of mean sq., V^ 2 , 
Vmean square. In England the initials R.M.S. are fre- 
quently used for root mean square. 

_ Effective E. M. F. . .. , , 

The ratio — — — is called the form-factor, 

Average E.M.F. J J 



since its value depends upon 
the shape of the pressure wave. 
For the curve Fig. 7, the form- 
factor is unity. As a curve be- 
comes more peaked, its form- Fi «- ?• 
factor increases, due to the superior weight of the squares 
of the longer ordinates. 

In the sinusoid the values found above give 

Form-factor = = i.n. 

2 -E m 

IT 



IO 



ALTERNATING-CURRENT MACHINES. 



6. Form Factor of Non-Sine Curves. — For the deter- 
mination of the form factor, three methods may be used, 
according to the character of the wave shape. First, if the 
equation of the curve is known, the analytical method may 
be employed. For example, take the ellipse, Fig. 8. Its 

b / o 

equation is y = - v 2 ax — or. 
a 




Fig. 8. 



The average ordinate is 



J y ax - I 

*/o aJ 



V2 ax — x 2 dx 



bVx — a / — — ~ , a 2 

- V 2 ax — x z + — vers 

a\_ 2 2 



aj a \2 ) 



ab 

7Z 7t 2 

and since a = - this becomes — • The square of the mean 
ordinate is 

Jy 2 dx — I (2 ax — x 2 ) dx — ax 2 — — 
a 2 J a 2 \_ 3 Jo 



b 2 I 3 8 a 3 \ 

« V 3/4 



.j/r 



71 3 7T 

7T 2 b^ 

but a = - , hence this becomes - — and it follows that the 



effective value is V/ — b. 
V 3 



PROPERTIES OF ALTERNATING CURRENTS. II 



Therefore the form factor = 



V% b W: 



1.04. 



4 
Second, the geometrical method may be used in calculating 

the form factor of simple wave shapes, as for example, Fig. 9. 



The average ordinate 

The effective value = y - 
The volume of Fig. 10 is 



area of Fig. 9 b \a -f- 2 a] 



base 



4 a 



olume o f Fig. 10 
base line 



2 ati 



ah 



b 2 (2 a + f a) = I ab\ 





*U ia ' J 



Fig. 9. 

Hence the effective value is 



Fig. 10. 



* 4 a 



4V: 



= b Vf and the form factor is , 3 = J — — = 1.09. 

And third, the form factor of irregular curves, as for 
example the lower E.M.F. curve of Fig. 5, may be deter- 
mined by the use of a planimeter. The average value 

= area of Flg< 5 = .60 E m . To obtain the effective value, 
base 

a curve of squared ordinates must be plotted. The area of 
this curve divided by its base is the mean ordinate and the 
square root of this mean square is the effective value of the 
voltage, which for the curve in question is .685 E m . Hence 

the form factor = '—— - =1.14. 
.00 



12 



ALTERNATING-CURRENT MACHINES. 



Probably no alternators give sine waves, but they ap- 
proach it so nearly that the value 1.11 can be used in most 
calculations without sensible error. 



IN PHASE 




Fig. 11. 



7. Phase The curves of the pressure and the current 

in a circuit can be plotted together, with their respective 
ordinates and common abscissae, as in Fig. 11. In some 

cases the zero and the 
maximum values of the 
current curve will occur 
at the same abscissae as 
do those values of the 
pressure curve, as in Fig. 
11. In such a case the 
current is said to be in phase with the pressure. In other 
cases the current will reach a maximum or a zero value at 
a time later than the corresponding values of the pressure, 
and since the abscissae are indifferently time or degrees, 
the condition is represented in Fig. 12. In such a case, 
the current is said to be out of phase with, and to lag be- 
hind the pressure. In 

still other cases the ,, t - 

curves are placed as in 
Fig. 13, and the current 
and pressure are again 
out of phase, but the 
current is said to lead Fig - I2 - 

the pressure. The distance between the zero ordinate of 
one sine curve and the corresponding zero ordinate of 
another, may be measured in degrees, and is called the 
angular displacement or phase difference. This angle of 
lag or of lead is usually represented by <f>. When one 




PROPERTIES OF ALTERNATING CURRENTS. 13 



LEADING CURRENT 



curve has its zero ordinate coincident with the maximum 
ordinate of the other, as in Fig. 14, there is a displacement 
of a quarter cycle (<j> = 90 ), and the curves are said to be 
at right angles. This 
term owes its origin to 
the fact that the radii 
whose projections will 
trace these curves, as 
in § 3, are at right 
angles to each other. Flg - I3, 

If the zero ordinates of the two curves coincide, but the 
positive maximum of one coincides with the negative maxi- 
mum of the other, as in 




RIGHT ANGLES 




Fig. 15, then <£ 



180 , 

op- 



Fig. 14. 

and the current 



and the curves are m 
posite phase. 

An alternator arranged 
to give a single pressure 
wave to a two-wire circuit is 
said to be a single phaser, 



in 



OPPOSITE PHASE 



the circuit a single-phase current. 
Some machines are arranged to give pressure to two dis- 
tinct circuits — each of 
which, considered alone, 
is a single-phase circuit 
— but the time of maxi- 
mum pressure in one is 
the time of zero pres- 
sure in the other, so 
that simultaneous pres- 
sure curves from the 
Fig. 16 




Fig. 15- 

two circuits take the form of 
Such is said to be a two-phase or quarter-phase 



14 



ALTERNATING-CURRENT MACHINES. 




system, and the generator is a two-phaser. A three-phase 
system theoretically has three circuits of two wires each. 
The maximum positive pressure on any circuit is displaced 
from that of either of the other circuits by 1 20 . As the 

algebraic sum of the cur- 
rents in all these circuits 
(if balanced) is at every in- 
stant equal to zero, the 
three return wires, one on 
each circuit, may be dis- 
pensed with, leaving but 
three wires. The three sim- 
ultaneous curves of E.M.F. 
are shown in Fig. 17. The term polyphase applies to any 
system of two or more phases. An /z-phase system has n 
circuits and n pressures with successive phase differences 

of - — degrees. 
n 

8. Power in Alternating-Current Circuits. — With a direct- 
current circuit, the power in the circuit is equal to the 
product of the pressure in volts by the current strength in 
amperes. In an alternating- 
current circuit, the instan- 
taneous power is the product 
of the instantaneous values 
of current strength and 
pressure. If the current 
and pressure are out of 
phase there will be some 
instants when the pressure will have a positive value and 
the current a negative value or vice versa. At such times 
the instantaneous power will be a negative quantity, i.e., 




PROPERTIES OF ALTERNATING CURRENTS. 15 




power is being returned to the generator by the disappear- 
ing magnetic field which had been previously produced by 
the current. This condition is shown in Fig. 18, where 
the power curve has for its ordinates the product of the 
corresponding ordinates of pressure and current. These 
are reduced by multiplying by a constant so as to make 
them of convenient size. 
The circuit, therefore, 
receives power from the 
generator and gives power 
back again in alternating 
pulsations having twice 
the frequency of the gen- 
erator. It is clear that 
the relative magnitudes Flg * l8 ' 

of the negative and positive lobes of the power curve will 
vary for different values of <f>, even though the original 
curves maintain the same size and shape. So it follows 
that the power in an alternating-current circuit is not 
merely a function of E and /, as in direct-current circuits, 
but is a function of E, I, and <£, and the relation is deduced 
as follows : — 

Let the accent (') denote instantaneous values. If the 
current lag by the angle <j>, then from § 3, 

E r = E m sin a, 
where, for convenience, 

a = 2 7r/?, 

and I'=I m si n ( a — <£)• 

Remembering that 

■p T 

E = — ^, and / = — £ (§ 5) the instantaneous power, 

V2 



P'=ET 



2 EI sin a sin (a — <£). 



16 

But 



ALTERNATING-CURRENT MACHINES. 



sin (a — cf>) = sin a cos <£ — cos a sin <£, 
so P' = 2 EI (sit? a cos <j> — sin a cos a sin <f>). 

Remembering that </S is a constant, the average power 
over 180 , 

„ 2^/ cos <f> C n ■ o , 2£/smd>r*. 

P = I Sin J cu/a — ■ I Sin a COS atf a 

TT Jo IT Jo 

2 EI cos <i> Yi i . I 77 2 EI sin cbYi . „ l 77 
= -a sm 2 a Sin 2 a . 

7T |_2 4 Jo 7T |_2 J 

■/* = i?/ COS <£. 

Should the current /^z<f the pressure by <j>°, then the 
leading equation would be 

P r = 2 EI sin a sin (a -f- </>), 

which gives the same expression, 

P = EI cos <£, 

which is the general expression for power in an alternating- 
current circuit. 

The above may also be shown by the use of vectors. Let 

OA and OB of Fig. 19 rep- 
resent the effective values of 
E.M.F. and current respect- 
ively, taking the former as the 
datum line and assuming the 
latter to lag <j> degrees behind 
the E.M.F. The line, OB, 
may be resolved into two com- 
ponents, one along OA and the other at right angles to it. 
These components, OP and OQ, are termed respectively 
the power and wattless components of the current. The 
actual power expended in the circuit is OA X OP = EI cos <}> 




Fig. 19. 



PROPERTIES OF ALTERNATING CURRENTS. 17 

and the wattless power, or that alternately supplied to and 
received from the circuit, is OA X OQ = EI sin <j>. 

Since, to get the true power in the circuit, the apparent 
power, or volt-amperes, must be multiplied by cos <£, this 
quantity is called the power factor of the circuit. If the 
pressure and current are in phase, <j> = o°, and the power 
factor is unity. 

It is important, at this point, to consider the graphical 
method of addition or subtraction of vector quantities, a 
process which is frequently employed in the treatment of 
alternating-current circuits. Let A and B, Fig. 20, be two 
lines whose lengths and whose directions respectively repre- 
sent the magnitudes and time or space locations of two vector 



/ 

/ 
{ - B 


\ 




/ 


/ a 


\ B / 
> / 



Fig. 20. 

quantities. These maybe E.M.F.'s, or currents as the case 
may be. The sum of two vectors is given in magnitude and 
in direction by the concurrent diagonal of a parallelogram 
the adjacent sides of which represent the vectors in size and 
direction. To substract one quantity from another vec- 
torially, it is but necessary to change its sign and add it to 
the other. Representing vectorial addition by the symbol 
© , and vectorial subtraction by © , the results of the various 
additions and subtractions of A and B become evident from 
the figure. 



i8 



ALTERNATING-CURRENT MACHINES. 



9. Non-Sine Waves. — As sine waves of E.M.F. or cur- 
rent are seldom obtained in practice, it is convenient, in 
accurate calculations, to refer to their equivalent sine waves. 
An equivalent sine wave is one having the same frequency 
and the same mean effective value as the given wave. Con- 
sider two non-sine waves, one of E.M.F. and the other of 
current, their zero or maximum values being displaced by an 
angle <f> n . The phase difference of these two non-sine waves 
cannot be considered as the angle cf> n , but is that phase dis- 
placement of their equivalent sine waves which would give 
the same average of the instantaneous power values as the 




non-sine waves. Therefore, to find the phase displacement of 
two non-sine waves, it is necessary, first, to plot a power curve 
and determine its average ordinate, P av ; second, to calculate 
the mean effective values of the curves, represented respec- 
tively by E and /; and third, to substitute these values in 
the equation P av = EI cos <£ from which <f> can be obtained. 
In general, it can be said, that when the form factors of 
both waves are less than that of the sine curve, their phase 



PROPERTIES OF ALTERNATING CURRENTS. 19 

difference is greater than the displacement of their zero or 
maximum values; and likewise, if their form factors exceed 
the value 1.11, then the phase displacement is less than the 
displacement of corresponding values of the curves. 

As a numerical example: Find the phase displacement 
of two semi-circular waves, having their zero values one- 
twelfth of a cycle or 30 apart. Let one be a pressure curve, 
whose maximum value is 120 volts, and the other, a current 
curve, whose maximum value is 6 amperes. These are 
shown in Fig. 21. 

The average ordinate of the power curve, determined by 
subtracting the negative area from the positive and then 
dividing by the base, is found to be 383 watts. The 



yf y 2 dx 
TZ 



effective value of each curve is »/ ^-° yj £ or tne c j rc } e 

V TZ 

being 2 bx — x 2 where b is the maximum ordinate. 
x dx — I x*dx biz 2 

Jo 3 , TZ 2 

U TZ 3 

7Z 71? 

Since b = - , this becomes — , and hence the effective 
2 6 

value is —-r * From the relations 
V6 

TZ TZ T , TZ 7Z _, 

- : —7- =6:1 and - : —_ = 120 : E, 
2 V6 2 V6 

it follows that the effective values of current and voltage 
are respectively 4.9 amperes and 98 volts. Then 383 = 98 X 
4.9 cos cj> y which gives as the phase displacement, <f> = 37°, 
instead of 30 . 



20 



ALTERNATING-CURRENT MACHINES. 



10. E.M.F.'s in Series. — Alternating E.M.F.'s that 
may be put in series may differ in magnitude, in fre- 
quency, in phase relation, 
and in form or shape of 
wave. 

If two harmonic E.M.F.'s 
of the same frequency and 
phase be in series, the re- 
sulting E.M.F. is merely 
the sum of the separate 
E.M.F.'s. This condition is 
shown in Fig. 22, in which 
thr two E.M.F.'s are plotted 
together, and the resulting E. vI.F. plotted by making its 
instantaneous values equal to the sum of the correspond- 
ing instantaneous values of the component E.M.F.'s. The 
maximum of the resultant E.M.F. is evidently 




Fig. 22. 



and since 



F m = E 1 

-a 



£1 



VI 



and 



£0 = 



V2 



Ex + F 2 



as was stated. 

If two E.M.F.'s of the same frequency, but exactly 
opposite in phase, be placed in series, it may be similarly 
shown that the resultant E.M.F. is the numerical differ- 
ence of the component E.M.F.'s. This case may occur in 
the operation of motors. 

The most general case that occurs is that of a number 
of alternating E.M.F.'p of the same frequency, but of 



PROPERTIES OF ALTERNATING CURRENTS. 21 




different magnitudes and phase displacements. The 
changes in magnitude and phase and the phase relation of 
the resulting curve of E.M.F. are shown in Fig. 23, where 
recourse is had once again to the harmonic shadowgraph. 
But two components, E x and E 2 , are treated, whose phase 
displacement is <f> v The radii vectors E im and E 2m are 
laid off from o with the proper angle <f> l between them, 
and the shadows traced by their extremities are shown in 
the dotted curves. The instantaneous value. of the result- 
ant E.M.F. is the algebraic sum of the corresponding in- 

A / / o/ j J7T\ \ Vff 1^ 27 T 

) ui r~mFz!n ut 90 \ \ iso\ s^y / y$60" 

/ 
f 

\«' 
v 



Fig. 23. 

stantaneous values of the component E.M.F.'s, and the 
resultant curve of E.M.F. is traced in the figure by the 
solid line. But this solid curve is also the trace of the ex- 
tremity of the line E m , which is the vector sum (the result- 
ant of the force polygon) of the component pressures, E im 
and E 21)l . This is evident from the fact that any instan- 
taneous value of the resultant pressure curve is the sum of 
the corresponding instantaneous values of the component 
curves, or (§ 3) 

E r = E lm sin at + E 2m sin (at -f- ^j). 
Again from the force polygon 

E m sin (o>/ -f- cf>) = E lm sin w/ -f- E 2m sin (w/ -f- <£i). 




22 



ALTERNATING-CURRENT MACHINES. 



Hence at any instant 

E' — E m sin (at + <j>), 

wherefore the extremity of the line E m traces the curve of 
resultant pressure, <£ being its angular displacement from 
E . If a third component E.M.F. is to be added in series, 
it may be combined with the resultant of the first two in 
an exactly similar manner. 

So it may be stated as a general proposition, that if any 
number of harmonic E.M.F.' s, of the same frequency, but 
of various magnitudes and 



E- 2 

Fig. 24. 



"E- 3 — -*| 



phase displacements, be 

connected in series, the 

resulting harmonic E.M.F. 

will be given in magnitude 

and phase by the vector sum of the component E.M.F. 's. 

The analytic expressions for E and <£ may be derived by 

inspection of the diagram, and are 

E = Vfi^sin^ + ^sin^oH-. • •] 2 -|-[ J £ , 1 cos</> 1 + J £ , 2 cos<£ 2 -j-- • -] 2 , 

and ^ sin & + F 2 sin cf> 2 + • • • t 

^ E x cos <£ x + E 2 cos </> 2 -j- . • • 




Fig. 25. 



As a numerical example, suppose three alternators, Fig. 
24, to be connected in series. Suppose these to give sine 
waves of pressure of values E x — 70, E % = 60, and E g = 40 



PROPERTIES OF ALTERNATING CURRENTS 23 

volts respectively. Considering the phase of E t to be 
the datum phase, let the phase displacements be <f> Y = o°, 
cf> 2 = 40 , and <j) 3 = 75 , respectively. It is required to find 
E and cf>. Completing the parallelograms or completing 
the force polygon as shown in Fig. 25, it is found that 
E = 148.7 volts and cf> = 32.1 . 

Alternating E.M.F.'s of different frequencies in series 
will give, in general, an irregular wave form. In practice, 
the frequencies of some E.M.F.'s are multiples of the fre- 
quency of another, called the fundamental E.M.F., or first 
harmonic. The pressure curve having twice this frequency 
is termed the second harmonic; another having three times 
this frequency, the third harmonic, and so on. The result- 
ant instantaneous E.M.F. is obtained by adding the pressure 




Fig. 26. 

values of all the components at that instant. It is expressed 

as £' = E lm sin wt + E 2m sin (2 at + 0J 

+ E 3m sin (scot + cf> 2 ) +... 
+ E nm sin (nut + <^_i) 



24 ALTERNATING-CURRENT MACHINES. 

where <j> v <f> 2 , . . . , <j> n -i, are the phase differences between 
E lm and E 2m , E lm and E 3m , . . . , E lm and E nm respectively 
when sin cot = o. 

When both odd and even harmonics are present, the 
resulting curve will have unlike lobes, but when only odd 
harmonics occur, as is usual in electrical machinery, the 
lobe above the horizontal axis and the other below it will 
be similar. Fig. 26 shows the resulting E.M.F. of three 
harmonic components for the values, E lm = 100 volts, 
E 2m = 40 volts, E 3m = 20 volts, <f> 1 = 30 and <£ 2 = 45 . 

Let it be required to find £', 2J seconds after the beginning 
of a cycle, the frequency of E x being 25 ^. 

cot = 2 iz 25 . 2\ = n6f 71 or — • 

A 71 J/ , , , \ 87T7T 1 7T 

2 cot = ^— and (2 cot + cp ) = — + - = i2 — • 
3 662 

2 7T 7T 

3 &>/ = 3 . - — ■ = 2 7r or o, then (3 cot + <£ 2 ) = - • 

tt 7-/ . 2 7T , . 3 It , . 7T 

Hence £ = 100 sin — ■ + 40 sm — + 20 sin — 
3 2 4 

= 86.6 — 40 + 14.1 = 60.7 

volts, which agrees with the value of the ordinate at 120 , 
in the figure. 

When the resulting pressure curve is given, it is possible, 
by graphical and analytical methods, to determine which 
harmonics are present, their maximum values, and phase 
displacements. 



PROBLEMS. 



PROBLEMS. 



i. What must be the speed of a 12-pole alternator to yield an 
E.M.F. of 60 cycles ? 

2. Find the instantaneous current value in a circuit, in which a 
25 ~w alternating current of 70.7 amperes flows, 6.0066 seconds after the 
completion of a cycle. 

3. How many amperes flow in a circuit, when the instantaneous 
value of the current is 5 amperes, 30 after the beginning of a cycle? 

4. What is the frequency of an E.M.F. which assumes its effective 
value every .01 second ? 

5. Find the average and effective values of a semi-circular wave 

shape, the maximum value being — ' Determine form factor. 

2 

6. Find the form factor of a triangular wave shape. 

7. Determine the form factor of the upper curve of Fig. 5. 

8. Find the instantaneous voltage produced by a 50 ^~ alternator, 
generating parabolic waves of 120 volts maximum value, if seconds 
after the beginning of a cycle. 

9. What is the phase displacement between E and /, respectively 
of 100 volts and 10 amperes maximum value, when the power in the 
circuit is 424 watts ? 

10. Find the phase difference of two non-sine waves of voltage and 
current, whose zero values are 45 apart. Let the wave shape be as 
shown in Fig. 9 and let E m = 100 volts and I m = 10 amperes. 

11. Four 60 -w alternators, generating respectively 100, 80, 90 and 
50 volts, are connected to a circuit. What will be the value of the result- 
ing pressure, and what will be its phase with respect to that of the 100 
volts, if the phase difference between successive components is 45 ? 

12. If the three E.M.F.'s, shown in Fig. 26, are impressed upon a 
circuit, what will be the resulting instantaneous voltage 5.015 seconds 
after the beginning of a cycle? 



26 ALTERNATING-CURRENT MACHINES. 



CHAPTER II. 

SELF-INDUCTION. 

ii. Self-Inductance. — The subject of inductance was 
briefly treated of in § 15, vol. i., of this work ; but, since it 
is an essential part of alternating-current phenomena, it 
will be discussed more fully in this chapter. When lines 
of force are cut by a conductor an E.M.F. is generated in 
that conductor (§ 13, vol. i.). A conductor carrying cur- 
rent is encircled by lines of force. When the current is 
first started in such a conductor, these lines of force must 
be established. In establishing itself, each line is con- 
sidered as having cut the conductor, or, what is equivalent, 
been cut by the conductor. This notion of lines of force 
is a convenient fiction, designed to render an understand- 
ing of the subject more easy. To account for the E.M.F. 
of self-induction, the encircling lines must be considered 
as cutting the conductor which carries the current that 
establishes them, during their establishment. It may be 
considered that they start from the axis of the conductor 
at the moment of starting the current in the circuit ; that 
they grow in diameter while the current is increasing ; that 
they shrink in diameter when the current is decreasing; 
and that all their diameters reduce to zero upon stopping 
the current. At any given current strength the conductor 
is surrounded by many circular lines, the circles having 
various diameters. Upon decreasing the strength those of 



SELF-INDUCTION. 2J 

smaller diameter cut the conductor and disappear into a 
point on the axis of the conductor previous to the cutting 
by those of larger diameter. The number of lines accom- 
panying a large current is greater than the number accom- 
panying a smaller current. 

The E.M.F. of self-induction is always a counter E.M.F. 
By this is meant that its direction is such as to tend to 
prevent the change of current which causes it. When the 
current is started the self -induced pressure tends to oppose 
the flow of the current and prevents its reaching its full 
value immediately. When the circuit is interrupted the 
E.M.F. of self-induction tends to keep the current flowing 
in the same direction that it had originally. 

12. Unit of Self-Inductancs. — The self-inductance, or 
the coefficient of self-induction of a circuit generally rep- 
resented by L or I, is that constant by which the time 
rate of change of the current in a circuit must be multi- 
plied in order to give the E.M.F. induced in that circuit. 
Its absolute value is numerically equal to the number of 
lines of force linked with the circuit, per absolute unit of 
current in the circuit, as is shown below. By linkages, or 
number of lines linked with a circuit, is meant the sum 
of the number of lines surrounding each portion of the 
circuit. For instance, a coil of wire consisting of ten 
turns, and threaded completely through by twelve lines 
of force, is said to have 1 20 linkages. 

The absolute unit of self-inductance is too small for 
ordinary purposes, and a practical unit, the henry, is used. 
This is io 9 times as large as the c. g. s. or absolute unit. 

The Paris electrical congress of 1900 adopted as the 
unit of magnetic flux the maxwell, and of flux density the 



28 ALTERNATING-CURRENT MACHINES. 

gauss. A maxwell is one line of force. A gauss is one 
line of force per square centimeter. If a core of an electro- 
magnet has a transverse cross-section of 30 sq. cm., and is 
uniformly permeated with 60,000 lines of force, such a 
core may be said to have a flux of 60,000 maxwells and a 
flux density of 2000 gausses. 

In § 13, vol. 1., it has been shown that the pressure gene- 
rated in a coil of wire when it is cut by lines of force is 

ndQ 

where n is the number of turns in a coil, and where e is 
measured in c. g. s. units, $ in maxwells, and t in seconds. 
In a simple case of self-induction the maxwells set up are 
due solely to the current in the conductor. Now let K be 
a constant, dependent upon the permeability of the mag- 
netic circuit, such that it represents the number of max- 
wells set up per unit current in the electric circuit ; then, 
indicating instantaneous values by prime accents, 

*' = Ki r , 
and d® = Kdi. 

The E.M.F. of self-induction may then be written 

di 

By the definition of the coefficient of self-induction, 
whose c. g. s. value is represented by /, 

di 

e ° = - l Jt' 

From the last two equations, it is seen that / = Kn. Kn is 
evidently the number of linkages per absolute unit current. 
The negative sign indicates that the pressure is counter 
E.M.F. 



SELF-INDUCTION. 29 

In practical units, 

B. — 1M. 

dt 

A circuit having an inductance of one henry will have a 
pressure of one volt induced in it by a uniform change of 
current of one ampere per second. 

13. Practical Values of Inductances. — To give the 
student an idea of the values of self-inductance met with in 
practice, a number of examples are here cited. 

A pair of copper line wires, say a telephone pole line, 
will have from two to four milhenrys (.002 to .004 henrys) 
per mile, according to the distance between them, the 
larger value being for the greater distance. 

The secondary of an induction coil giving a 2" spark has 
a resistance of about 6000 ohms and 50 henrys. 

The secondary of a much larger coil has 30,000 ohms 
and about 2000 henrys. 

A telephone call bell with about 75 ohms has 1.5 henrys. 

A coil found very useful in illustrative and quantitive 
experiments in the alternating-current laboratory is of the 
following dimensions. It is wound on a pasteboard cylinder 
with wooden ends, making a spool 8.5 inches long and 2 
inches internal diameter. This is wound to a depth of 1.5 
inch with No. 16 B. and S. double cotton-covered copper 
wire, there being about 3000 turns in all. A bundle of 
iron wires, 16 inches long, fits loosely in the hole of the 
spool. The resistance of the coil is 10 ohms, and its in- 
ductance without the core is 0.2 henry. With the iron 
core in place and a current of about 0.2 ampere, the induc- 
tance is about 1.75 henrys. This coil is referred to again 
in § 16. 



30 ALTERNATING-CURRENT MACHINES. 

The inductance of a spool on the field frame of a gene- 
rator is numerically 

T _ ® H 



io 8 // 



where <£ is the total flux from one pole, n the number 
of turns per spool, and I f the field current of the machine. 
It is evident that the value of L may vary through a wide 
range with different machines. 

14. Things Which Influence the Magnitude of L. — If all 

the conditions remain constant, save those under considera- 
tion, then the self-inductance of a coil will vary : directly as 
the square of the number of turns ; directly as the linear 
dimension if the coil changes its size without changing its 
shape ; and inversely as the reluctance of the magnetic 
circuit. 

Any of the above relations is apparent from the follow- 
ing equations. The numerical value of the self-induc- 
tance is 

/= n — • 
1 

As shown in Chapter 2, vol. i., 

_ M.M.F. _ 4 imi 
reluctance c 

where c is the mean length in centimeters of the magnetic 
circuit, A its mean cross-sectional area in square centi 
meters, and fx is permeability. 

Then, if (R stand for the reluctance, 

n 4 -mi .A 4 irn 2 

I = - • = 4 7J7T u — = ^ * 

1 c * ^ c (ft 

fiA 

which is independent of i. 



SELF-INDUCTION. 31 

If, as is generally the case, there is iron in the magnetic 
circuit, it is practically impossible to keep ^ constant if any 
of the conditions are altered ; and it is to be particularly 
noted, that with iron in the magnetic circuit, L is by no 
means independent of /. 

15. Formulae for Calculating Inductances. — Circle: 
For a cylindrical conductor of radius r cm. and length / cm., 
bent into a circle and surrounded with a medium of unit 
permeability the self-inductance in henries is 

L — IO -9 2 / \ log e - — I.508 > • 

This is accurate to within 0.2 % when the radius of the circle 
is greater than ten times that of the cylindrical conductor. 

Straight Wire: For a straight cylindrical conductor of 
radius r cm. and length I cm. in a medium of unit permea- 
bility the self-inductance in henries is 



10- 9 2 / 



log e ^- -0.75 



Parallel Wires: For a return circuit of two parallel cylin- 
drical wires of radius r cm., d cm. apart from center to center, 
each of permeability \i and of I cm. length, the self-inductance 
in henries is 



L = 10 9 4 / 



1 d m" 
log 6 - + - 



Solenoids: The formula given below for the self-induct- 
ance of a solenoid of any number of layers will give results 



32 ALTERNATING-CURRENT MACHINES. 

accurate to within one-half of a per cent even for short 
solenoids, where the length is only twice the diameter, the 
accuracy increasing as the length increases. 

l\/ 4 a 2 + / 2 3 * J 
j[(m - i) ^ 2 + (m - 2) < + . . . ][Va* + P - J a J 

+ i[m(w— i)a 1 2 + (m— i)(w— 2)a 2 2 + (w— 2)(m— 3)a 3 2 + . . .] 

LvV + / 2 JJ 

Where 

w is the number of layers, 

a is the mean radius of the solenoid, 

a lf a 2 , a 3 , . . . a m are the mean radii of the various 

layers, 
/ is the length of the solenoid, 
da is the radial distance between two consecutive 

layers, 
n is the number of turns per unit length. 

A simple and convenient formula for the calculation of 
the self -inductance of a single-layer solenoid is as follows: 



L = 4* 2 



V4 a 2 + I 2 3 * 



where a is the mean radius and / is the length of the 
solenoid. 



SELF-INDUCTION. 33 

The natural logarithms used in preceding formulae can be 
obtained by multiplying the common logarithm of the num- 
ber, the mantissa and characteristic being included, by 2.3026. 

The inductance of all circuits is somewhat less for 
extremely high frequencies than for low ones. 

16. Growth of Current in an Inductive Circuit. — If a 
constant E.M.F. be applied to the terminals of a circuit 
having both resistance and inductance, the current does 
not instantly assume its full ultimate value, but logarith- 
mically increases to that value. 

At the instant of closing the circuit there is no current 
flowing. Let time be reckoned from this instant. At 
any subsequent instant, / seconds later, the impressed 
E.M.F. may be considered as the sum of two parts, E 1 
and E r . The first, E v is that part which is opposed to, 
and just neutralizes, the E.M.F. of self-induction, so that 
E l =-E s ; 

but Es = ~ L W 

r dI 

The second part, E n is that which is necessary to send 
current through the resistance of the circuit, according to 
Ohm's Law, so that 

E r = RI. 

If the impressed E.M.F. 

E=E r +E 1 =RI + L d ^, 

then (E - EI) dt = Idl, 
and dt = — =-= dl = 



E - RI R E-RI 



34 



ALTERNATING-CURRENT MACHINES. 



Integrating from the initial conditions t = o, f=o to any 
conditions t = t, /=/'. 



L 
R 

Rt_ 



\[og(E-RI r )-\o Z E\ 



-*<cr) 



and 



r 



E E _*t 
R~R* L 



E 
R 



I — e L 



) 



where e is the base of the natural system of logarithms. 

This equation shows that the rise of current in such a 
circuit is along a logarithmic curve, as stated, and that when 



t is of sufficient magnitude to 

R . 



negli- 




render the term e~Z 

gible the current will follow 

Ohm's Law, a condition that 

agrees with observed facts. 

Fig. 27 shows the curve of 

growth of current in the coil 

referred to in §13. The curve 

is calculated by the above formula for the conditions 

noted. 

The ratio — is called the time constant of the circuit, 
R 

for the greater this ratio is, the longer it takes the current 
to obtain its full ultimate value. 



.02 .03 .04 .05 .06 .07 

SECONDS 

Fig. 27. 



17. Decay of Current in an Inductive Circuit. — If a cur- 
rent be flowing in a circuit containing inductance and re- 
sistance, and the supply of E.M.F. be discontinued, 
without, however, interrupting the continuity of the circuit, 
the current will not cease instantly, but the E.M.F. of 



SELF-INDUCTION. 35 

self-induction will keep it flowing for a time, with values 
decreasing according to a logarithmic law. 

An expression for the value of this current at any time, 
/ seconds after cutting off the source of impressed 
E.M.E, may be obtained as in the preceding section. Let 
time be reckoned from the instant of interruption of the 
impressed E.M.E. The current at this instant may be 



represented by 
induction. 


E 


and is 


due solely to tl 


Therefore 




E = 


dt 


or 




RI = 


dt 




• 


\ dt = 


L dl 
R I 



E 
Integrating from the initial conditions t = o, I = — , to 

the conditions, t = t> I = /', 





\ DECAYING CURRENT 


\ E.M.F.-O 


\ R — 10 


\ L-.2 


V- '- ,0 


;" 



r* l r*'di 

J/^-rJe^- 



t = 



L 
R 


log 


I' 
E 
R 


E _ 
— e 


-f' 


j 



.0.1 .02 .03 .0* .05 ,06 .07 ,08 . 09 . 1 j j-f 

seconds ana i 

Fig. 28. 

which is seen to be the term that had to be subtracted in 
the formula for growth of current. This shows clearly 
that while self-induction prevents the instantaneous attain- 
ment of the normal value of current, there is eventually no 
loss of energy, since what is subtracted from the growing 
current is given back to the decaying current. 

Fig. 28 is the curve of decay of current in the same cir- 



36 ALTER'NATING-CURRENT MACHINES. 

cuit as was considered in Fig. 27. The ordinates of the 
one figure are seen to be complementary to those of the 
other. 

18. Magnetic Energy of a Started Current. — If a cur- 
rent / is flowing under the pressure of E volts, the power 
expenditure is EI watts, and the work performed in the 
interval of time dt is 

dW=EIdt. 

During the time required to establish a steady flow of 
current after closing the circuit, the impressed E may be 
considered as made up of two parts, one, E r , required to send 
F through the resistance of the circuit, and the other, E s , 
which opposes the E.M.F. of self-induction. E r I dt appears 
as heat, while E S I dt is stored in the magnetic field. Since 

Es ~ L dt' 
dW=-LIdL 

Integrating through the full range, from o to W and from 
o to /, 



dW = -L / Idl. 
J 

.-. W = - \LI\ 



which is an expression for the work done upon the magnetic 
field in starting the current. When the current is stopped 
the work is done by the field, and the energy is returned to 
the circuit. 

The formula assumes the value of L to be constant during 
the rise and fall of the current, but this is not the case with 
an iron magnetic circuit. If L is taken as the average of 



SELF-INDUCTION. 37 

the instantaneous values of self-inductance between the 
limiting values of the current, then the formula for the 
energy stored in the field still holds true. 

Since iron has always a hysteretic loss, some of the 
energy is consumed, and the work given back at the dis- 
appearance of the field is less than that used to establish 
the field by the amount consumed in hysteresis. 

19. Current Produced by a Harmonic E.M.F. in a Cir- 
cuit Having Resistance and Inductance. — Given a circuit 
of resistance R and inductance L upon which is impressed 
a harmonic E.M.F. E of frequency f, to find the current 
I in that circuit. 

Represent by w the quantity 2-nf. 

At any instant of time, /, let the instantaneous value of 
the current be /'. 

To maintain this current requires an E.M.F. whose value 
at this instant is I r R. Represent this by E ' r . 

From § 3, in a harmonic current, 

I f = I m sin (at, 
hence, E/ = RI m sin <o/. 

Evidently E 7 ! has its maximum value RI m = E rm at (nt = go° 
or 270 , and its effective value is E r = RI. 

The counter E.M.F. of self-induction at the same instant 
of time, /, is 

But as before, /'= I m sin <ot, 

so dl r = (i)I m cos (u/ dt, 

and El = — mLI m cos wt. 



38 ALTERNATING-CURRENT MACHINES, 

Evidently E, has a maximum value of — a>L/ m = E m at 
tot = o° or 180 , and its effective value 

E s = - toZI. 

It is clear that the impressed E.M.F. must be of such 
a value as to neutralize E s and also supply E r . But these 
two pressures cannot be simply 
added, since the maximum value of 
one occurs at the zero value of the 
other; that is, they are at right 
angles to each other, as defined in 
§ 7. Reference to Fig. 29 will 
make it clear that combining these at right angles will 
give as a resultant the pressure ^/E r 2 + E s 2 ; and it is this 
pressure that the impressed E.M.F. E must equal and 
oppose. So 




from which 



e = V(/#) 2 + (vLiy, 

1= , E • 

\ll? + co 2 Z 2 



This is a formula which must be used in place of Ohm's 
Law when treating inductive circuits carrying harmonic 
currents. It is evident that, if the inductance or the fre- 
quency be negligibly small (direct current has / = o), the 
formula reduces to Ohm's Law ; but for any sensible val- 
ues of co and L the current in the circuit will be less than 
that called for by Ohm's Law. 

The expression \lR 2 -f w 2 Z 2 is called the impedance of 
the circuit, and also the apparent resistance. The term R 
is of course called resistance, while the term <oZ, which is 
2 vfLy is called the reactance. Both are measured in ohms. 

The effective value of the counter E.M.F. of self-indue- 



SELF-INDUCTION. 39 

tion can be determined as follows, without employing the 
calculus ; that it must be combined at right angles with 
RI is not directly evident. Disregarding the direction of 
flow, an alternating current i reaches a maximum value i m , 
2f times per second. The maximum number of lines of 
force linked with the circuit on each of these occasions is 
li m . The interval of time, from when the current is zero 
with no linkages, to when the current is a maximum with 

li m linkages, is — - - second. The average rate of cutting 
4/ 

lines, then, is — - , and is equal to the average E.M.F. of 

4? 
self-induction during the interval. It has the same value 

during succeeding equal intervals ; i.e., 

e,av = = - 4.A»- 

4? 

The effective value is (§ 5) therefore, 
e s = — 2 itfli = <iili, 
and in practical units, 

E, = - 2 tt/LL 

Since the squares of the quantities R, L, and w enter 
into the expression for the impedance, if one, say R, is 
moderately small when compared with L or a), its square 
will be negligibly small when compared with L 2 or w 2 . The 
frequency, because it is a part of o>, may be a considerable 
factor in determining the impedance of a circuit. 

Having recourse once again to the harmonic shadow- 
graph described in § 3, the phase relation between im- 
pressed E.M.F. and current may be made plain. It has 
already been shown that E r and E s are at right angles to 



40 



ALTERNATING-CURRENT MACHINES. 



each other. Since the pressure E r is the part of the im- 
pressed E.M.F. which sends the current, the current must 
be in phase with it. Therefore there is always a phase 
displacement of 90 between / and E s . This relation is 
also evident from a consideration of the fact that when i" 
reaches its maximum value it has, for the instant, no rate 
of change; hence the flux, which is in phase with the cur- 
rent, is not changing, and consequently the E.M.F. of self- 
induction must be, for the instant, zero. That is, / is maxi- 
mum when E s is zero, which means a displacement of 90 . 
In Fig. 30 the triangle of E.M.F.'s of Fig. 29 is altered 




Fig. 30. 



to the corresponding parallelogram of E.M.FSs, and the 
maximum values substituted for the effective. If now the 
parallelogram revolve about the center o, the traces of 
the harmonic shadows of the extremities of E m , E rm and E sm 
will develop as shown. It is evident that the curve E/ — 
and so also the curve of current — ■ leads the curve E f by 
the angle <f>. It is clear that the magnitude of cf> depends 
upon the relative values of L and R in the circuit, the exact 
relation being derived from the triangle of forces. 

. E s coLT ojL 2 TtjL 
tan <£ = — s = = - — = — '— • 

v E, RI R R 



SELF-INDUCTION. 41 

Furthermore 

COS (f> = -±, 

that is, the cosine of the angle of lag is equal to the ratio 
of the volts actually engaged in sending current to the 
volts impressed on the circuit, and this ratio is again equal 
to the power-factor as stated in § 8. 

20. Instantaneous Current produced by a Harmonic 
E.M.F. in a Circuit having Resistance and Inductance. — 

The E.M.F. at any instant /, impressed upon a circuit con- 
taining resistance and inductance, must be of such magni- 
tude as to send the instantaneous current F through the 
resistance, and also to neutralize the E.M.F. of self-induc- 
tion. That is 

dT f 

E f = RF + Z^f • 
dt 



But E' = E m sin cot. (Art. 3.) 

dF 
dt 1 



Hence E m sin cot = RF + L 



dl R -r, E m . 

fi dt 
Multiply by integrating factor e 

dt L L 



//? 7?/ 

— dt — — and writing in differential form, 

dFe L + Fe L y dt = ^f sin cot. e L dt. 



42 ALTERNATING-CURRENT MACHINES. 

The second term is in the form da x = a x log e a dx. 

m . m x p rj_ 

Hence dI'e L + I'd [e L \ = ^ sin cot . e L dt. 

Since the first member is in the form d(xy) = y dx + x dy, 
it may be replaced by d iFe 1 }. Integrating, 

Rt 27 r* Rt 

Fe' 1 = ^ J e 1 'sin wt dt + C. 



__Rt p s* iu ^m 

Hence F = e L ^ J e L sincotdt + Ce L . (i) 



Rt ^ n Rt m 

L 

To determine value of the integral, use the formula 

I u dv = uv — f v du. 
Let u = sin cot, then du = co cos cot dt, and let 

Rt j Rt -g 

dv = e L dt = ke L ^dt. 



R L 



Hence v = - e L • 

R 

Then the integral becomes 

e L sin cot dt = — e L sin cot — / e L cos cot dt. 

Use the same formula again for this second integral, but here 
u = cos cot, hence du = — co sin w/ <i/; and where 

Rt 

dv =e L dt. 



SELF-INDUCTION. 



43 



Hence 

Then 



/ 



T — 
Rt j Rt 

e L sin cot dt = —e L sin cot 
R 

Rt 



ojV t , 



R 2 



Rt 

e L cos cot 



R 2 



s 



e sin cot dt. 



(■ + f)/« ? 



Or i + 



sin cot at = — e sin w/ — — -e cos w/. 



2? 



# 2 



Substitute value of integral in (i), then 



r = e 






' T Rt 7 , iJC 

^ T • , <^ " X 

— £ sin &tf — — — e cos cot 
R R^ 

R 2 + co 2 L 2 



R 2 



+ Ce 



R 



sin co — co cos &>/ 



+ Ce 



(*) 



Let the angle cfi be chosen so that tan <f> = — , thus repre- 

senting the angle of lag of the current behind the E.M.F. 
Therefore 

* R 

COS = — - , 



and 



sin = 



Vi? 2 + co 2 L 2 

coL 
VR 2 + a/Z 2 



Hence 12 = V^+^Z 2 cos and co= lA_+^!^l£i 



44 ALTERNATING-CURRENT MACHINES. 

Substitute these values in the numerator of (2 ). Then 

/' = f?x 



VR 2 + co 2 L 2 cos <f> sin cot-\/R 2 + co 2 L 2 sin <j> cos cot ] -§, 

R 2 +co 2 L 2 



E m ~ -5 

Vi? 2 + a/Z, 2 



Or /' = / m o (cos0 sin co/- sin cos^Q + Ce l 



And //= /p 2r2 sin (** " #) + C« ' ■ (3) 

_m 
The term Ce L shows the natural rise of the current when 
the voltage is first impressed upon the circuit. After a few 
cycles have been completed this term may be neglected. 

Then V = Em sin (cot - 0). (4) 

VR 2 + co 2 L 2 

This expression gives the instantaneous current in a cir- 
cuit having resistance and inductance at any instant, when 
a harmonic E.M.F. is impressed upon that circuit. When 
L = o and <p—o, the equation reduces to 

V = I m sin cot as in Article 3. 

21. Choke Coils. — The term choke coil is applied to 
any device designed to utilize counter electromotive force 
of self-induction to cut down the flow of current in an 
alternating-current circuit. Disregarding losses by hyster- 
esis, a choke coil does not absorb any power, except that 
which is due to the current passing through its resistance. 
It can therefore be more economically used than a rheostat 
which would perform the same functions. 



SELF-INDUCTION. 45 

These coils are often used on alternating-current circuits 
in such places as resistances are used on direct-current 
circuits. For instance, in the starting devices employed in 
connection with alternating-current motors, the counter 
E.M.F. of inductance is made to cut down the pressure 
applied at the motor terminals. The starter for direct- 
current motors employs resistance. 

It is often desirable to adjust the reactance of choke coils, 
and for this purpose several simple arrangements may be 
utilized. The coil may have a sliding iron core, or its wind- 
ing may have several taps. Choke coils having U-shaped 
magnetic circuits are sometimes provided with movable 
polepieces, which serve to change the length of the air gap. 

Since a lightning discharge is oscillatory in character and 
of enormous frequency, a coil which would offer a negligible 
impedance to an ordinary alternating current will offer a 
high impedance to a lightning discharge. This fact is recog- 
nized in the construction of lightning arresters. A choke 
coil of but few turns will offer so great an impedance to a 
lightning discharge that the high-tension, high-frequency 
current will find an easier path to the ground through an 
air gap suitably provided than through the machinery, and 
the latter is thus protected. 

A choke coil for this purpose has no iron core, and con- 
sists of a few turns of wire, insulated from one another, 
wound in spiral or helical form. A lightning arrester choke 
coil used in railway service, for station use, is shown in 
Fig. 31. 

The choking effect is not alone due to the high impedance 
offered to an oscillatory discharge, but also due to the "skin 
effect" of the wire. By this is meant, the tendency of the 
alternating current to have a greater density near the surface 



46 ALTERNATING-CURRENT MACHINES. 

than along the axis of the conductor, thus increasing the 
resistance. To illustrate, a ij" round copper conductor 
offers a true resistance, twice as great as its ohmic resistance, 
to a 130^ alternating current. Even in small wires, the 
true resistance presented to currents of very high frequency, 




Fig. 31. 

such as those produced by wireless telegraph transmitters, 
greatly exceeds the ohmic resistance, and therefore con- 
ductors are required possessing a large surface compared 
to the cross-section. 

Choke coils are also used in connection with alternating- 
current incandescent lamps, to vary the current passing 
through them, and in consequence to vary the brilliancy. 



PROBLEMS. 47 



PROBLEMS. 



i . What is the field winding inductance of a bi-polar generator, having 
7500 ampere turns per spool and a total flux of 2.4 mega-maxwells, 
when the exciting current is 2 amperes? 

2. Find the inductance of a cast steel test ring coil of 300 turns when 
carrying 6 amperes, the test ring being 6" outside and 5" inside diameter 
and i\" in axial depth. 

3. Determine the inductance of a pole-line 10 miles long and consist- 
ing of a pair of No. 1 copper wires separated by a distance between 
centers of 24 inches. 

4. Determine the self-inductance of a solenoid consisting of 10 layers 
of No. 16 double-cotton covered wire, 100 turns per layer, wound upon 
a cylindrical wooden core 2 inches in diameter. 

5. Find the value of the current in a circuit having 5 ohms resistance 
and an inductance of 0.15 henry, .03 seconds after impressing no volts 
upon that circuit. 

6. What is the time constant of a circuit in which the current reaches 
half of its ultimate value .0018 second after connection with a source of 
E.M.F.? 

7. What would be the current .02 second after suppressing the E.M.F. 
in the circuit of problem 5, a constant flow having been previously 
established ? 

8. Determine the energy stored in the magnetic field of the generator 
of problem 1, assuming L to be constant during the rise or fall of the 
current. 

9. Find the current produced by a 60 ~ alternating E.M.F. of 120 
volts in a circuit having 10 ohms resistance and an inductance of .04 
henry. What is the power factor of the circuit ? 

10. What should the inductance of the circuit of problem 9 be, to 
attain a power factor of 85%? 

n. Derive an expression for the current in a circuit whose resistance 
and reactance are equal. What will be the power factor? 

12. Find the instantaneous value of a 25 ^~ alternating current, 
2.342 seconds after impressing a harmonic E.M.F. of 125 volts maxi- 
mum upon a circuit which has a resistance of 8 ohms and an induct- 
ance of 0.04 henry. 



48 ALTERNATING-CURRENT MACHINES. 



CHAPTER III. 

CAPACITY. 

22. Condensers. — Any two conductors separated by a 
dielectric constitute a condenser. In practice the word is 
generally applied to a collection of thin sheets of metal 
separated by thin sheets of dielectric, every alternate metal 
plate being connected to one terminal and the intervening 
plates to the other terminal. The Leyden jar is also a 
common form of condenser. 

The function of a condenser is to store electrical energy 
by utilizing the principle of electrostatic induction. When- 
ever a difference of potential is impressed upon the con- 
denser terminals, stresses are set up in the dielectric which 
exhibit themselves electrically as a counter electromotive 
force, opposing and neutralizing the impressed E.M.F. 
During the period of establishment of the stresses a current 
flows through the dielectric, and it is known as a displace- 
ment current. This, however, ceases to flow as soon as 
the counter electromotive force of dielectric polarization is 
equal in magnitude to the impressed E.M.F. The con- 
denser is then said to be charged. It should be remem- 
bered that the charge resides in the dielectric as the result 
of the stresses produced in it by the impressed E.M.F. 

The nature of the stresses in the dielectric can be more 
readily understood by considering the conductors to be 
surrounded by an electrostatic field. This field may be 
considered as composed of electrostatic lines of force, shown 



CAPACITY. 49 

in Fig. 32, which indicate by their directions the directions 
of the stresses, and by their nearness to each other the 
magnitude of the stresses. The greater the impressed 
E.M.F., the greater will be the number of these lines and 

\ \ \ \ \ 1 / / / y ^ 

— - -\ N V V/// 

Oxv\\\ I U/.'/s' 



J 1 / ,'s-z^\ ' n 1 1 1 1 HI Li f I I fV^V^v \ \ 

! 1 'f'ffli ' I ' l ih l lll Xhlli | l ! l, l , l il!, i ^ ,V\ \ \ 




Fig. 32. 

the greater will be the charge. The property of a dielectric 
which opposes the passage of this dielectric flux may be 
termed its obstructance, and it is similar in this respect to 
reluctance in opposing the passage of magnetic flux, and to 
resistance in opposing the flow of current. The obstruc- 
tivity of a dielectric is three hundred times the reciprocal 
of its specific inductive capacity or its dielectric constant, 
which is the ratio of the electric strain to the stresses pro- 
duced by it in the dielectric. 

No dielectric is capable of supplying more than a definite 
maximum amount of counter electromotive force per unit 
of length measured along a line of force. If the impressed 
potential difference exceeds this maximum counter E.M.F., 
which is the measure of its dielectric strength, the dielectric 
is ruptured and breaks down mechanically. Of course, if 



5o 



ALTERNATING-CURRENT MACHINES. 



the dielectric is a liquid or a gas, it will be restored to its 
original state when the impressed E.M.F. is diminished. 
At the point of rupture, a current in the form of a spark or 
an arc passes from one conductor to the other, the tendency 
being to lessen the potential difference. Such rupture, 
followed by an arc, is a frequent source of trouble in electric 
machinery. 

The dielectric strength of any dielectric depends upon 
its thickness, the form of the opposed conducting surfaces, 
and the manner in which the E.M.F. is applied, whether 
gradually, suddenly, or periodically varying. It has been 
stated that the dielectric strength approximately varies 
inversely as the cube root of the thickness, showing that a 
thin sheet is relatively stronger than a thick one of the same 
material. For example, the dielectric strength of crystal 
glass when 5 mm. thick is 183 kilo volts per centimeter, but 
when 1 mm. thick it is 285 kilovolts per centimeter. In the 
following table giving the dielectric strengths of various 
materials, the particular thicknesses for which the values 
are given are stated: 



Material. 


Thickness in mm. 


Dielectric Strength 
in Kilovolts per cm. 






Air 


IO 


29.8 


Air 


I 


43- 6 


Glass 


5 


183 


Mica 


1 


610 


Mica 


O.I 


1150 


Micanite 


1 


400 


Linseed Oil 


6 


84 


Vaseline Oil 


6 


60 


Lubricating Oil 


6 


48 


Ebonite 


2 


43° 



The capacity of a condenser is numerically equal to the 
quantity of electricity with which it must be charged in 



CAPACITY. 51 

order to raise the potential difference between its terminals 
from zero to unity. 

If the quantity and potential be measured in c. G. s. 
units, the capacity, c, will be in c. G. s. units. If practical 
units be employed, the capacity, c, is expressed in farads. 
The farad is the practical unit of capacity. A condenser 
whose potential is raised one volt by a charge of one 
coulomb has one farad capacity. The farad is io -9 times 
the absolute unit, and even then is too large to conven- 
iently express the magnitudes encountered in practice. 
The term microfarad (toouooo farad) is in most general 
use. 

In electrostatics, both air and glass are used as dielec- 
trics in condensers; but the mechanical difficulties of con- 
struction necessitate a low capacity per unit volume, and 
therefore render these substances impracticable in electro- 
dynamic engineering. Mica, although it is expensive and 
difficult of manipulation, is generally used as the dielectric 
in standard condensers and in those which are intended 
to withstand high voltages. Many commercial condensers 
are made from sheets of tinfoil, alternating with slightly 
larger sheets of paraffined paper. Though not so good as 
mica, paraffin will make a good dielectric if properly 
treated. It is essential that all the moisture be expelled 
from the paraffin when employed in a condenser. If it 
is not, the water particles are alternately attracted and 
repelled by the changes of potential on the contiguous 
plates, till, by a purely mechanical action, a hole is worn 
completely through the dielectric, and the whole condenser 
rendered useless by short-circuit. Ordinary paper almost 
invariably contains small particles of metal, which become 
detached from the calendar rolls used in manufacture. 



52 ALTERNATING-CURRENT MACHINES. 

These occasion short-circuits even when the paper is 
doubled. 

The capacity of a condenser is proportional directly to 
the area and inversely to the thickness of the dielectric. 
It is also directly proportional to the dielectric constant of 
the insulating material, which, in addition to the definition 
already cited, may be defined as the number expressing the 
ratio of increase of the capacity of an air condenser, when 
the air is entirely replaced by that dielectric. This constant, 
usually represented by K, decreases with the temperature 
and with the time of charge. For these reasons the values 
of K given by different observers differ considerably, but 
some accepted values are given in the following table: 



DIELECTRIC 


CONSTANTS 


AT 15 


C. 




Flint Glass (dense) 


IO.I 


Quartz 






4.55 


Flint Glass (light) 


6 -57 


Sulphur 






2.9 to 4.0 


Crown Glass (hard) 


6.96 


Shellac 






2.7 to 3.0 


Mica 


6.64 


Ebonite 






2-05 to 3.15 


Tourmaline 


6.05 


Paraffin 


Wax 




2.0 to 2.3 



The resistance of a condenser is not infinite, but a meas- 
urable quantity, and is usually expressed in megohms per 
microfarad, or, when referring to cables, in megohms per 
mile. Hence there is always a leakage from one charged 
plate to the other, both through the dielectric and over its 
surface. Poor insulation may occasion a considerable loss 
of energy appearing in the form of heat, and is therefore 
to be avoided. 

Analogous to magnetic hysteresis in iron, is dielectric 
hysteresis in condensers, but, contrary to the former, it 
decreases as the frequency increases. Thus, at a frequency 
of the order of 10 million cycles, dielectric hysteresis is 
entirely absent. A dielectric having a high hysteretic con- 



CAPACITY. 53 

stant, such as glass — 6.1, may consume a considerable 
amount of energy on low frequency circuits, this loss also 
appearing as heat. 

23. Capacity Formulae. — The following formulae, in 
which r is the radius of the conductor and I its length, both 
in centimeters, give the capacity in microfarads of con- 
ductors with respect to the earth: 

Sphere in free space, 



900,000 
Circular disk in free space, 

r 



C = 



i,4i3>7 2 ° 



One cylindrical wire in free space, 

c- L_ 



4,144,680 log 10 - 
r 

One cylindrical wire h cm. from the earth, 

4,144,680 l0g 1Q y- • 

In the following formulae, giving the capacity of con- 
densers of various forms, only that portion of the dielectric 
flux which passes perpendicularly between the conducting 
surfaces is considered; that is, the end flux shown by the 
curved dotted lines in Fig. 32 is neglected. Under this 
consideration, the following expressions may only be used 
when the thickness of the dielectric is very small compared 
to the conductor area. 



54 ALTERNATING-CURRENT MACHINES. 

Two concentric spheres, 

Y Y K 

q _ i_2^ where y > r v 

900,000 (^-rj 

Two concentric cylinders, 

C = where r 9 > r.. 

4,144,680 log 10 ^ 

Two cylindrical wires d cm. apart, 

C= »- 



8,289,360 log 10 - 



Two circular plates, d cm. apart, 

r 2 £ 



C = 



3,600,000 d 

From this last formula, another may be readily derived 
for the calculation of the capacity of a condenser having n 
dielectric sheets, and having its symbols expressed in inches. 
The capacity is 

^ An „ 

C = .00022=; — K, 

where A is the area of each sheet in square inches, and t is 
its thickness in mils. 

The following data of a condenser, used in duplex 
telegraphy, give an idea of capacity and dielectric resistance. 

The condenser consists of tinfoil and paper sheets, the 
former being brought out alternately to one terminal and 
then to the other. There are 92 sheets of beeswaxed paper, 
7X5 inches and two mils thick, which constitute the dielec- 
tric. The capacity of the condenser is 1.47 microfarads, 
and its dielectric resistance is 160 megohms. 



CAPACITY. 55 

24. Connection of Condensers in Parallel and in Series. 

— Condensers may be connected in parallel as in Fig. 33. 
If the capacities of the individual Cl 

condensers be respectively C\, C 2 , C 3 , 
etc., the capacity C of the combina- 
tion will be — 



c 3 



Fig- 33- 



C = C t + C 2 + C 3 + 

For the potential difference on each 

condenser is the same, and equal to 

the impressed E.M.F., and the total charge is equal to the 

sum of the individual charges, or 





E = 


E t 


= E, 


= £3 = 






and 


e- 


Q, 


+ Q2 


+ Q3 + 






Then by division 


Q 

E 


E, 


E 2 


+%+ 






But by definition 


Q 

E 


c, 


£1 


= C t and 


so 


on, 


therefore 


C 


= 


c t + c 2 + c 3 


+ 


. . . 



The parallel arrangement of several condensers is equiva- 
lent to increasing the number of plates in one condenser. 
An increase in the number of plates results in an increase in 
the quantity of electricity necessary to raise the potential 
difference between the terminals of the condenser one volt; 
that is, an increase in the capacity results. 

If the condensers be connected in series, as in Fig. 34, 
the capacity of the combination will be 

c= l - • 



c c c 

^1 ^2 b 3 



56 



ALTERNATING-CURRENT MACHINES. 



For, if a quantity of positive electricity, Q, flow into the 
left side of C v it will induce and keep bound an equal neg- 
ative quantity on the right side of C v and will repel an 
equal positive quantity. This last quantity will constitute 
the charge for the 
left side of C a . — 



Ci 



E T *~< E- s 



-E-3-— 



The operation is 

repeated in the j* — e~ 

case of each of Fig - 34 * 

the condensers. It is thus clear that the quantity of 
charge in each condenser is Q. The impressed E.M.F. 
must consist of the sum of the potential differences on the 
separate condensers. Let these differences be respectively 
E lf E 2 , E g) etc. Then the impressed E.M.F. 







E = £ 1 + E % + E s + • • • • 


But 


a 


Q p Q p -Q etc 

= ~r> ^2 = t^' -^3—7^' eic -> 
Cj_ c 2 c 3 


and also, 




F Q 


therefore 




Q = Q + Q = Q + 

C- C]_ C 2 c 3 

r z 






I I I 

c ~*~ ~r ~c 

'-'l ^2 ^3 



As an example, consider three condensers of respective 
capacities of 1, 2, and 5 microfarads. Since the factor to 
reduce to farads will appear on both sides of the equations, 
it may here be omitted. With the three in multiple (Fig. 
33), the capacity of the combination will be 

C = 1 + 2 + 5 = 8 mf . 



CAPACITY. 



5/ 



With the three in series (Fig. 34), 



C = 



1 2 5 



= .588 mf. 



With the two smaller in parallel and in series with the 
larger (Fig. 35), 



C 



1 1 

--7— + - 
1 + 2 5 



1.875 mf: 



d c 2 

c, | H 1 f^ 



Fig. 35- 



c 3 
Fig. 36 



With the two smaller in series and in parallel with the 
larger (Fig. 36), 



C 



1 2 



+ 5 = 5.666 mf. 



= C 



If with any condensers 

Cj = C 2 = c 3 = • 
then, with ;/ in multiple, 

C = nC lt 

and with ft in series, 

C = - C v 

n 

It is interesting to note that the formulas for capacities 
in parallel and in series respectively are just the reverse of 
those for resistances in parallel and in series respectively. 

25. Decay of Current in a Condensive Circuit. — The 

opposition to a flow of current which is caused by a con- 



58 ALTERNATING-CURRENT MACHINES. 

denser is quite different from that which is caused by a 
resistance. To be sure, there is some resistance in the 
leads and condenser plates, but this is generally so small 
as to be negligible. The practically infinite resistance of 
the condenser dielectric does not obstruct the current as 
an ordinary resistance is generally considered to do. The 
dielectric is the- seat of a polarization E.M.F. which is de- 
veloped by the condenser charge and which grows with it. 
It is a counter E.M.F. ; and when it reaches a value equal 
to that of the impressed voltage, the charging current is 
forced to cease. 

To find the current at any instant of time, t, in a circuit 
(Fig. 37) containing a resistance R and a capacity C, the 
constant impressed pressure E must 
be considered as consisting of two 
variable parts, one E r , being active 
in sending current through the re- 
sistance, and the other part, E c , 



I'M- 



being required to balance the po- Fig. 37. 

tential of the condenser. Then at 

all times 

e' = e; + EJ. 

Let time be reckoned from the instant the pressure E is 

E 
applied ; when, therefore, t = o and / = — . Consider the 

current at any instant of time to be /'. Then if it flow 
for dt seconds it will cause dQ coulombs to traverse the 
circuit, and 



1'=?$- or dQ = Tdt, 
at 



from which 



</= Cl'dt. 



By definition, 
therefore, 



CAPACITY. 

c-V- 

, v I 1 ' 



dt 



E = 



C C 

And by Ohm's Law, 

e/ = rr, 

so at this instant of time 






whence 



E'c=xcr+ Ci'dt, 

which upon differentiating, becomes 

o = RCdl' + l'dt. 
Integrating f dt = _ RC C" ^ 

Jo J/, I 



CONDENSER 

CHARGING CURRENT 

E-=100 V. 

C = 2 MF. 
=.000002 F. 




Solving for I' 9 



59 



Fig. 38. 



r-'r-k. 



which is the expression 
sought. Like the corre- 
sponding expression for an 
inductive circuit, it is loga- 
rithmic. 



Fig. 38 is a curve showing the decay of current in a 
condenser for the conditions indicated. The product RC 



60 ALTERNATING-CURRENT MACHINES. 



is the time constant of a condensive circuit and is similar 

to the ratio — in an inductive circuit. 
R 

26. Energy Stored in Dielectric. — A current I flowing 
in a condensive circuit against a dielectric polarization of 
E volts, represents a power of EI watts. The work per- 
formed in an interval of time dt is 

dW = Eldt 

and represents the elementary work done in establishing 
the stresses in the dielectric. Since 

E = 2 an d 7 dt = dQ y 



there results by substitution 

which when integrated through the full range, that is 

o 2 

becomes W = -^~ joules. 

2 C 

This is the expression for the energy required to 
establish the dielectric stresses when the current is first 
applied, and also the expression for the energy returned 
to the circuit by the dielectric when the impressed E.M.F. 
is withdrawn. 

27. Condensers in Alternating-Current Circuits — Hy- 
draulic Analogy. — Imagine a circuit consisting of a pipe 
through which water is made to flow, first one way, then 



CAPACITY. 6l 

the other, by a piston oscillated pump-like in one section 
of it. The pipe circuit corresponds to an electric circuit, 
the pump to a generator of alternating E.M.F., and the 
flow of water to a flow of alternating current. Further 
imagine one section of the pipe to be enlarged, and in it 
placed a transverse elastic diaphragm. This section cor- 
responds to a condenser. Its capacity with a unit pressure 
of water on one side depends upon the area of the dia- 
phragm, its thinness, and the elastic coefficients of the 
material of which it is made. In a condenser the capacity 
depends upon the area of the dielectric under strain, its 
thinness, and the specific inductive capacity of the dielec- 
tric employed. As the water surges to and fro in the 
pipe, some work must be done upon the diaphragm, since 
it is not perfectly elastic. This loss corresponds to the 
loss in a condenser by dielectric hysteresis. The fact that 
the diaphragm is not absolutely impervious to water cor- 
responds to the fact that a dielectric is not an absolute 
electric insulator. As the diaphragm may be burst by too 
great a hydrostatic pressure, so may the dielectric be rup- 
tured by too great an electric pressure. 

28. Phase Relations. — To understand the relation be- 
tween pressure and current in a condensive circuit, con- 
sider the above analogy. Imagine the diaphragm in its 
medial position, with equal volumes of water on either side 
of it, and the piston in the middle of its travel. This 
middle point corresponds to zero pressure. When the pis- 
ton is completely depressed there is a maximum negative 
pressure, when completely elevated, a maximum positive 
pressure, if pressure and flow upward be considered in the 
positive direction. If the piston oscillate in its path with a 



62 



ALTERNATING-CURRENT MACHINES. 




regular motion, it is clear that the water will flow upward 
from the extreme lowest to the extreme highest position of 
the piston. That is, there will be flow in the positive direc- 
tion from the maximum negative to the maximum positive 
values of pressure. The direction of 
flow is seen to remain unchanged while 
the piston passes through its middle 
position or the point of zero pressure. 
These facts are indicated in Fig. 39, 
which shows that portion of the pipe 
having the piston and the diaphragm. 

Returning to electric phenomena, if a 
harmonic E.M.F. be impressed upon 
any circuit, a harmonic current will flow 
in it. So in a circuit containing a con- 
denser and subject to a sinusoidal 
E.M.F., the current flow will be sinusoidal. This flow 
will be in the positive direction from the negative maximum 
to the positive maximum of pressure, and in a negative 
direction from the positive 
maximum to the negative 
maximum, as described 
above. This necessitates 
that the zero values of 
current occur at the maxi- 
mum values of pressure; Flg ' 4 °' 
and since the curves are both sinusoids, their relation 
may be plotted as in Fig. 40. It is immediately seen that 
these curves are at right angles, as described in § 7, and that 
the current leads the pressure by 90 . 

Reference again to the hydraulic analogy will show that 
the condenser is completely charged at the instant of 




CAPACITY. 



63 



maximum positive pressure, discharged at the instant of 
zero pressure, charged in the opposite direction at the 
instant of maximum nega- 
tive pressure, and finally 
discharged at the instant 
of the next zero pressure. 
Thus the charge is zero 
at the maximum current 
flow, and at a maximum 
at zero current, that is, 
when the current turns 
and starts to flow out. 
Fig. 41. 




Fig. 4x. 

These points are marked in 



29. Current and Voltage Relations. — If a sinusoidal 
pressure E of frequency / be impressed upon a condenser, 



1 1 



the latter is charged in — - seconds, discharged in the next 
4/ 



4/ 



seconds, and charged and discharged in the opposite 



direction in the equal succeeding intervals. The maximum 
voltage E m = V2 E (§ 5), hence the quantity at full charge 
is 

Q m = V2EC. 

The quantity flowing through the circuit per second is 

4fQm = 4f\ / 2EC. 
This number therefore represents the average current, or 

I av = 4 V2JEC. 
From § 5, the effective current 

7Z 



I = 



V] 



6 4 



ALTERNATING-CURRENT MACHINES. 



whence 



7 = 2 nfCE, 



and 



E = 



7. 



2 7ZJC 

The last is an expression for the volts necessary to send 
the capacity current through a circuit. The expression 
1/2 71 fC is called the capacity reactance of the circuit. It is 
analogous to 2 njL, the inductive reactance of an inductive 
circuit. 

If the circuit contain both a resistance R and a capa- 
city C, the voltage E impressed upon it must be considered 
as made up of two parts, E r , which sends current through 
the resistance and is therefore in phase with the current, 
and E c , which balances the counter pressure of the con- 
denser and is therefore 90 behind the current in phase. 

By Ohm's Law 

E r = RI, 
and from above 

7. 



2 njC 



The impressed E must overcome the resultant of these 
two E.M.F.'s] and since they are at right angles 

E = VE? 



E r - RI 




or 



E 2 



E 



Fig. 42. 



^+(ik) 



The relation of the E.M.F.'s is shown graphically in 
Fig. 42, where the current, which is in phase with the 
pressure E r , is seen to lead the impressed pressure by the 
angle cf>. 



CAPACITY. 65 

30. Instantaneous Current in a Circuit Having Capacity 
and Resistance. — The value of the E.M.F., impressed 
upon a circuit containing capacity and resistance at any 
instant /, must be sufficient to send the instantaneous cur- 
rent I' through the resistance and also neutralize the E.M.F. 
of dielectric polarization. Hence 

E' = E r ' + E/. 



But 


E' = E m sin cot, 


(§3) 




E/ = I'R, 


(§25) 


and 


EJ - f™ 


(§25) 




C ' 




therefore 


Crdt 

E m sin cot = I'R A * 





c 

I'dt 
Differentiating coE m cos cot dt = R dl' ^ • 

rdt j_ 

Multiplying by integrating factor e^ RC or e RC and dividing 
by R, this becomes 

dFe hc + I'e^'f c = ^e* c cos a>t dt. 

The second term is in the form da x = a x log e a dx, hence 
this is 



(«*) 






Since the first member is in the form d(xy) = y dx + x dy it 
equals d \I'e RC ). Substitute and integrate, then 



<oE, 



- ^^ j e RC cos col dt + C, 



66 



ALTERNATING-CURRENT MACHINES. 



t T? f* t t 

or I' = e~ T ^ aL Y~ J e™ cos cot dt + Ce"^ . (i) 

To determine value of the integral, use the formula 

/ udv — uv — I vdu, where u = cos tot, hence du = — to 

— — dt 

sin cot dt; and where dv = e RC dt = RCe RC -zr^r, hence v = 

RC 

t 

RCe RC \ Then 

J e* 6 cos cot dt = RCe™ cos cot + RCco J e™ sin cot dt. 

The second integral is in the same form, but here u = sin 
cot, hence du = to cos cot dt, dv and v remain the same. Then 



/■ 



e R0 cos cot 



t t 

RCe RC cos tot + R 2 C 2 coe RC sin cot -R 2 CW 



e RC cos cot dt. 



[i + R 2 C 2 to 2 \ j e RC cos cot dt = RCe RC cos cot 

t 
+ R 2 C 2 coe RC sin cot. 

Substitute value of integral in (i), there results 

t t 

RCe RC cosojt + R 2 C 2 toe RC sin cot 



r = e AC 



R 



i + R 2 C 2 co 2 



or 



+ Ce RC . 

r = coCE 7 

T f = F 



f" cos COl 



cot + RCco sin cot' 



cos tot + R sin o>/ 



>C 



> 2 C 2 



+ Ce 



+ C6> 



#6' 



l 1 
RC 



(2) 



CAPACITY. 67 

Let the angle cf> be chosen so that 



tan <£ = ^£ = T 



R coCR 

thus representing the angle by which the current leads the 
E.M.F. Therefore 



and 



^-V^+=5?^*- 



CO C 

By substitution in (2), there results, 



v' 



R 2 + 



1 

2^-2 _ t m 

RC 



F = E m [sin <f> cos cot + cos <£ sin cot] + Ce 

co C 

or J' = Em - sin (W + <f>) + Ce" ** (3) 



vH^J 



where the exponential term shows the natural current decay 
in a condensive circuit when the E.M.F. is first applied. 
Neglecting this term, (3) reduces to 

sin {cot + <f>), (4) 



\ R 2 



>C 



giving an expression for the instantaneous current in a 
circuit, having resistance and capacity, at any instant when 
a harmonic E.M.F. is impressed upon that circuit. When 
the capacity of a circuit is an infinitesimal, such as is the case 
when its two terminals are slightly separated, then in the 



68 ALTERNATING-CURRENT MACHINES. 

formula, C = o and the current is also zero, which is 
evidently true for an open circuit. When the circuit con- 
tains no capacity relative to itself, and only resistance, then 

the term (—p^j should not enter the equation, which will 

then reduce to 

F = I m sin (cut + <j>), as in § 3. 

PROBLEMS. 

1. Determine the capacity of a pair of No. 000 line wires, two feet 
apart, and three miles long. 

2. Calculate the dielectric constant of the condenser mentioned in 
§ 23. What is its insulation resistance expressed in megohms per 
microfarad ? 

3. Derive the formula C = .000225 — ^ of § 23. 

t 

4. Find the equivalent capacity of the group of condensers shown in 






Fig. 43- 

Fig. 43, the number adjacent to each condenser representing its capacity 
in microfarads. 

5. If a constant E.M.F. of 150 volts is applied to the terminals A 
and B of the group of condensers shown in Fig. 43, what will be the 
voltage across the terminals of each condenser? 

6. If a circuit having a resistance of 10 ohms and a capacity of 20 
microfarads has a constant E.M.F. of 100 volts impressed upon it, how 
long will it take for the current to sink to half its initial value? 

7. Determine the energy which can be electrically stored in a cubic 
inch of mica dielectric when the applied potential is 450 volts per mil 
thickness. 



PROBLEMS. 69 

8. Find the current produced by a 25^ alternating E.M.F. of 100 
volts in a circuit having 25 ohms resistance and a capacity of 30 micro- 
farads. What is the power factor of the circuit? 

9. It is desired to construct a condenser of crown glass plates 10 X 12 
inches so that the power factor of its circuit having 12.5 ohms resistance 
shall be 90% for an oscillatory current of 80,000 cycles. How many 
plates will be required if the thickness of each is .15 inch? 

10. Determine the instantaneous value of a 60^ alternating current 
5.71 seconds after impressing a harmonic E.M.F. of 220 volts (effective) 
upon a circuit having a resistance of 100 ohms and a capacity of 25 
microfarads. 



7o 



ALTERNATING-CURRENT MACHINES. 



CHAPTER IV. 



ALTERNATING-CURRENT CIRCUITS. 



31. Resistance, Inductance and Capacity in an Alter- 
nating-Current Circuit. — In general, alternating-current 
circuits have resistance, inductance and capacity. An 
expression for the current flow in such a circuit may be 
derived mathematically, as in § 34, or the current may be 
found graphically by combining results already obtained. 
In § 19 it was shown that the counter E.M.F. due to the 
inductive reactance of a circuit is 2 TtjLI and leads the 
current by 90 , and in § 29 it was shown that the E.M.F. 





1 I 


\ 


(J 


I «T/0 


V" 






PHASE 



OP I 





f- . 


\ 


AG 

PHASE 


1 






r/C | 





Fig. 44, 



Fig. 45- 



of dielectric polarization due to the capacity reactance of a 
circuit is - — — and lags behind the current by 90 ; hence 

2 7T/C 

these two E.M.F.'s are opposite in phase, or 180 apart. 
These relations are shown in Fig. 44, where the inductive 
reactance is greater than that due to capacity, and in Fig. 
45, where the latter exceeds the former, the resistance being 
the same in both cases. The common factor / is omitted 



ALTERNATING-CURRENT CIRCUITS. Jl 

in these diagrams, as is very often done for convenience, 
but it should be remembered that neither resistance, reac- 
tance nor impedance is a vector quantity. Clearly the 
impedance resulting from the three factors, R, L and C, is 
represented in direction and in magnitude by the hypothe- 
nuse as shown, and the impressed pressure is / times this 
quantity. 

The general expression for the flow of an alternating 
current through any kind of a circuit is therefore 

E 



v^+R^-^/cJ 



the quantity within the brackets indicating an angle of lag 
of current when positive, and an angle of lead when negative. 

32. Definitions of Terms. — In considering the flow of 
alternating currents through series circuits and through 
parallel circuits, continual Use must be made of various 
expressions, some of which have been defined during the 
development of the previous chapters. For convenience 
the names of all the expressions connected with the general 
equation 

E 



I = 



v* 2+ R L -^y 



will be given and defined. 

I is the current flowing in the circuit. It is expressed 
in amperes, and lags behind or leads the pressure, by an 
angle whose value is 

2 kJL — — ^ 

jr , -1 2 7T/C 

<j> = tan * — • 

R 



72 ALTERNATING-CURRENT MACHINES. 

E is the harmonic pressure, of maximum value V2 E, 
which is applied to the circuit, and has a frequency /. It 
is expressed in volts. 

R is the resistance of the circuit, and is expressed in 
ohms. It is numerically equal to the product of the im- 
pedance by the cosine of <f>. 

L is the inductance of the circuit, and is expressed in 
henrys. 

C is the localized capacity of the circuit, and is expressed 
in farads. 

2 njL is the inductive reactance of the circuit, and is 

expressed in ohms. 

1 

2 7T/C 

is the capacity reactance, or capacitance, of the circuit, and 
is expressed in ohms. 

is the reactance of the circuit, and is expressed in ohms and 
usually represented byX. It is numerically equal to the 
product of the impedance by the sine of <f>. 



\/ R2 + [ 2 * IL ~ djcl or VR2 + x * 

is the impedance or apparent resistance of a circuit, and is 
expressed in ohms and usually represented by Z. 



1 1 

or z' 



V* + ["/^^feT 



jc. 

the reciprocal of the impedance, is the admittance of the 
circuit, and is represented by F. It is expressed in terms 



ALTERNATING-CURRENT CIRCUITS. 



73 



of a unit that has never been officially named, but which has 
sometimes been called the mho. There are two compo- 
nents of the admittance, as shown in Fig. 46. 

The conductance of a circuit, usually represented by g, 
is that quantity by which E must be multiplied to give the 





Fig. 46. 



component of / parallel to E. It is expressed in the same 
units as the admittance, and is numerically equal to 

COS (f> 

Z 



or Y cos (f> : 



but 



cos (j) = — , hence 

A/ 



Z 2 



The susceptance of a circuit, represented by b, is that 
quantity by which E must be multiplied to give the com- 
ponent of I perpendicular to E. It is measured in the same 
units as the admittance, and is numerically equal to 



sin 



but 



sin 



or Y sin <j>, 



X 1, a X 

— , hence h- — . 



Admittance may then be expressed as 



Y = Vg 2 + b 2 . 

It should be noticed that while admittance is the recip- 
rocal of impedance, conductance is not the reciprocal of 



74 



ALTERNATING-CURRENT MACHINES. 



resistance, nor is susceptance the reciprocal of reactance. 
This becomes evident, upon considering numerical values 
in connection with the impedance right-angled triangle, 
e.g. 3, 4 and 5 for the sides. 



33. Representation of Impedance and Admittance by 
Complex Numbers. — The problem of determining current, 
voltage and phase relations in alternating-current circuits 
may be solved graphically, by means of vector diagrams, 
or trigonometrically. To facilitate the solution of particular 
problems by the latter method, use is made of complex 
numbers. 

In Fig. 47, let / be the current produced in a circuit 

by the harmonic 
E.M.F., E, the cur- 
rent lagging behind 
the electromotive 
force by the angle <j>. 
Taking the rect- 
angular reference 
axes x and y as 
shown, both E and 
/ may be resolved 
into components along them. Let the symbol j be placed 
before the v-components, thus distinguishing them from the 
^-components. Then 

E = e± + je 2 




Fig. 47. 



and 



h + Pi 



the plus sign indicating vector addition at right angles of the 
x and y components respectively. But E may also be 
resolved into a component in phase with the current and 



ALTERNATING-CURRENT CIRCUITS. 75 

into another at right angles thereto, that is, it may be 
expressed as 

E = RI + jXI, 

and substituting the values of E and 7, there results 

e, + je 2 = Ri t + jRi 2 + jXi t + fXi 2 . 

Both RI and XI may be resolved into components along 
the axes of reference as indicated, and hence it follows that 

e i = Rh — Xi 2 

and e 2 = Ri 2 + Xi t . 

Then Ri x - Xi 2 + jRi 2 + jXi x = Ri, + jRi 2 + jXi x + fXi 2 . 

Therefore — Xi 2 = fXi 2 

or f = — 1 

and j = V — 1, 

which is therefore the interpretation of the symbol /, as 
already denned. 

From the foregoing, 



R +JX' 

but J=|, (§32) 

hence the impedance Z may be properly represented by 
R + jX. 

Admittance, being the reciprocal of impedance, may then 

be represented by — — , and multiplying both numerator 

and denominator by R — jX, there results, 

R - jX _ R- jX _ R-jX 

(R + jX) (R - jX) R 2 - fX 2 R 2 + X 2 ' 



?6 ALTERNATING-CURRENT MACHINES. 

Separating, Y = ^J— _ j _A__ = | . _ j j. . 

But — = g and y 2 = 6. (§32) 

Hence the admittance Y is to be represented by g — jb. 

34. Instantaneous Current in a Circuit Having Induct- 
ance, Capacity and Resistance. — In § 20 an expression was 
derived for the instantaneous current produced by a har- 
monic E.M.F. in a circuit having inductance and resistance, 
and in § 30 a similar expression was derived for a circuit 
having capacity and resistance. Proceeding along the same 
lines, a general expression could be obtained for the instan- 
taneous current produced by a harmonic E.M.F. in any 
alternating-current circuit, that is; in one having inductance, 
capacity and resistance. This method, however, is rather 
cumbersome, and a simpler one is given as follows: 

The harmonic E.M.F. is represented by E m e j(0t , an 
expression which results from the use of Maclaurin's Series, 
that is, 

/(*) =[/(*)L=o + *[/'(*)L=o 

L? Li 

where /' '(x), f '(x), /'"(#), .... are the respective deriva- 
tives of j(x). 

When the function is sin 6, 

. 6 s 6 5 6' 

sin = 6 — ._ + .-—_ -f . . . 

li Is Iz 

and when the function is cos 6, 

cos d = 1 — ,- + , ,7 + - • • 

2 4 |6 



ALTERNATING-CURRENT CIRCUITS. 77 

When, however, the function is z? e , then 

\l \i k I \i t I 

Remembering that f = — i, j 4 = i, f = — i, . . . this 
becomes 



1 -'-■ '-- : ' /; : r,+--] 



Hence e je ^ coz + j sin 0. 

Multiply through by E m and replace by cot, then 
E m ^ mt = E m (cos cot + j sin cot), 

which is evidently a proper expression for a harmonic E.M.F. 
Consider a circuit having a resistance R, a capacity C 
and an inductance L. The E.M.F. impressed upon this 
circuit must be of such magnitude as to neutralize both the 
counter E.M.F. of self-induction and the E.M.F. of di- 
electric polarization, and also send the instantaneous 
current V through the resistance. Therefore 

e' = e; + EJ + E/ 
or E m J" =L^+ ^j rdt + FR - ( J ) 

Since the current is of the same character as the impressed 
E.M.F., it may be represented by Be* **, where B is a 
constant to be determined. Then 





I 1 = Be?** 


and 


f-;w- 


and j Fdt -- 


= B fe^dt^—^ = ^e 

J ]CO i CO 



fat _ j ~ j(ot 

CO 



78 



ALTERNATING-CURRENT MACHINES. 



Substituting these values in (i), there results 

E m e^ = jwLBe*"' - y-^e** + RBth*, 

coC 



or 



E m = J5 



Hence 23 = 



[ R - + i( mL ~£ft m ' 



or 



23 = 



t + fa-zc) R -\" L ~ic) 
R *~? (" L ~ £ 



E. 



R 2 + Ll - 



["-'h-£il 



■c 



But F - Be 1 '"* = 5 (cos cot + y sin cot), 

and substituting value of B, there results 



/' = 



R2 + [» L -£ 

[cos &>£ + j sin &>/], 



r?[«-'(- 1 -^ 



or 

r = 



a 



— — (2? cos w/ + wL sin cot) 



. . R cos w/ +U>L 
+ y[ 2? sin a»/ — ojL cos cot) 



ALTERNATING-CURRENT CIRCUITS. 79 

Assuming the impressed harmonic E.M.F. as a simple sine 
function, then only the second part within the bracket of 
this expression need be taken; hence 

V = ^ \r sin cot - (ojL - ^Vos cot\. (2) 

r> + Ll-^-) 1 v wC/ j 

V cod 

Now let an angle (f> be chosen so that 



tan 6 = 



toL - — 

coC 

R~ 



then coL = V R 2 + ( coL — — - ) sin </> 

coC * V coCJ 

and R = \/ R 2 + LoL - ~) cos <£. 

Substituting these in (2), 

Em\l R 2 + (ojL - -^Y 
1 = — L sm &t cos 9 —cos cot sm 9 J 

or I' = £ " sin H - *), (3) 

which is the required expression for the instantaneous 
current produced by a harmonic E.M.F. in a circuit con- 
taining inductance, capacity and resistance. 

When L — o, the expression reduces to the form given 
in § 30; and when the circuit has no capacity with respect 

to itself, the term — drops out, and the expression reduces 
cots 

to the form given in § 20. It follows, then, that equation 
(3) may be applied to any alternating-current circuit. 



SO ALTERNATING-CURRENT MACHINES. 

35. Resonance. — An electrical circuit is said to be 
resonant, or in resonance with an impressed E.M.F., when 
the natural period of that circuit and the period of the 
E.M.F. are the same. The natural period of the circuit 
is the reciprocal of that frequency at which the current is a 
maximum. By reference to the formula 

E 



I = 



V R 2 + (2 njL - 



2 TtjCt 



it becomes evident that the maximum current is — , which 

R 

occurs when 

2 *}L ^— = o, 

2 7T/C 

that is, when the capacity and the inductance are so pro- 
portioned that their reactances are equal. From this 
relation, it follows that the critical frequency at which 
resonance occurs is 

1 -Mis _ 

and that the natural period of the circuit is 2 n \/LC. 

To show the current values for different frequencies, a 
curve as in Fig. 48 may be drawn. It is plotted for a series 
circuit, having 5 ohms resistance and an inductance of 
0.422 henry and a capacity of 6 microfarads; upon which 
circuit is impressed a harmonic E.M.F. of 100 volts. The 
frequency of the impressed E.M.F. required for resonance 
is seen to be 100 **. At this frequency the potential dif- 
ference between the terminals of the condenser = — 

2 71 fC 

= 5300 volts, and that across the inductance coil is 2 njLI 
— 53°o vo^s also, whereas only 100 volts is impressed upon 



ALTERNATING-CURRENT CIRCUITS. 



81 





1 


■JO 






-15 




CO 

U. 
EC 

a 
S 
< 


■10 


RE80NANCE 




CURVE 1 1 


~ii 



50 



FREQUENCY 



Fig. 48. 



150 



the circuit. Hence when resonance occurs m a circuit in 
which the capacity and the inductance are in series, the 
potential difference 
across either may 
rise to such a value 
as to puncture the 
insulation of the 
apparatus. If the 
capacity and induc- 
tance be in parallel, 
enormous currents 
may flow between 
the two. This is 
because the two are balanced, and the one is at any time 
ready to receive the energy given up by the other; and a 
surging once started between them receives periodical in- 
crements of energy from the line. This is analogous to 
the well-known mechanical phenomena that a number of 
gentle, but well-timed, mechanical impulses can set a very 
heavy suspended body into violent motion. The frequency 
of these impulses must correspond exactly to the natural 
period of oscillation of the body. In this parallel arrange- 
ment, serious damage is likely to result from resonance in 
overloading and burning out the conductors between the 
inductance and capacity. 

The protection of transformers and other station appara- 
tus against high-potential surges coming from transmission 
lines is effected by the use of choke coils interposed between 
the lines and the station wiring. It is essential for proper 
protection, that the electrostatic capacity of this wiring be 
as small as possible and that the choke coil have as low an 
inductance as will allow the lightning arrester to take up 



82 ALTERNATING-CURRENT MACHINES. 

the discharge, so that the frequency of resonance will be 
raised, thus decreasing the liability of picking up destructive 
voltages from line impulses or lightning discharges. 

36. Damped Oscillations. — When a condenser is dis- 
charged through a circuit having resistance and inductance, 
an oscillatory current flows, the maximum values of which 
decrease logarithmically. The ratio of two successive 

maximum values can be shown equal to £ 2 f, where a is the 

damping factor and is equal to ■ — - , R being the high-fre- 

2 -L/ 

OL 

quency resistance of the circuit. The entire exponent, — - 

2 / 

is called the logarithmic decrement, and is represented by d. 
Hence < 



R 
4 Lf 



and replacing / by — V 777" > (§ 35 ) 

2 71 T LC % 



there results 



4^- 



2 • L 
The effective current value of a train of damped oscilla- 




Fig. 49. 



tions, one of which is shown in Fig. 49, can be deduced by 
considering the energy, stored in the condenser at each 



ALTERNATING-CURRENT CIRCUITS. 83 

charge and discharged n times per second, to be consumed 
in heating the conductors. Then from § 26, 

PR = n%, 

2 C 

and substituting the value of R in the expression for 3, there 
is obtained 

4 CI- Y L 

and since Q = E m C, 

2 = riTcEJC ./C 
4 d h" 

Hence the effective value of the current is 



^ 



/ = 

2 T d' T£ 

The natural frequency of a circuit in which a decaying 
oscillatory current flows is dependent only upon R, C and 
L, and may be obtained from the formula 

/ = ~ V ~TF~ a2 where TF > a2 ' 

When a 2 > — — - , the current is unidirectional and decreases 
as in Fig. 38. 

37. Polygon of Impedances. — Consider a circuit having 
a number of pieces of apparatus in series, each of which 
may or may not possess resistance, inductance, and capacity. 
There can be but one current in that circuit when a pressure 



8 4 



ALTERNATING-CURRENT MACHINES. 



is applied, and that current must have the same phase 
throughout the circuit. The pressure at the terminals of 
the various pieces of apparatus, necessary to maintain 
through them this current, may, of course, be of different 
magnitude and in the same or different phases, being 
dependent upon the values of R, L, and C. Therefore to 
determine the pressure necessary to send a certain alter- 



R t L, C, 



WMH 




Fig. 50. 



nating current through such a series circuit, it is but neces- 
sary to add vectorially the pressures needed to send such a 
current through the separate parts of the circuit. This is 
readily done graphically although in many cases the various 
quantities may be of such widely different magnitudes that 
it will be found more convenient to make use of trigono- 
metrical expressions and methods. 

Fig. 50 shows the pressures (according to § 31) neces- 
sary to send the current I through several pieces of ap- 



ALTERNATING-CURRENT CIRCUITS. 85 

paratus, and the combination of these pressures into a 
polygon giving the resultant pressure E necessary to send 
the current / through the several pieces in series. In 
these diagrams, impedance is represented by the letter Z. 
C x and C 3 are localized, not distributed capacities. 

For practical purposes, the quantity I, which is common 
to each side of the triangle, may be omitted ; and merely 
the impedances may be added vectorially in a " polygon of 
impedances," giving an equivalent impedance, which, when 
multiplied by /, gives E. 

Inspection of the figure shows that the analytical ex- 
pression for the required E is 

£^(H 1+Jlt +. ■ ■f+[ m (Z 1 +l 2 +. ■ .)-(-L + -£, + . . .)] 

The pressure at the terminals of any single part of the 
circuit is 






E,= 

It is evident that 

E x + E 2 + > E, 

and it is found by experiment that the sum of the potential 
differences, as measured by a voltmeter, in the various 
parts of the circuit, is greater than the impressed pressure. 

38. A Numerical Example Applying to the Arrangement 
Shown in Fig. 50. — Suppose the pieces of apparatus to 
have the following constants : 



86 



ALTERNATING-CURRENT MACHINES. 



i?! = 85 ohms, L x = .25 henry, C x = .000018 farad (18 mf.) 

i? 2 = 40 ohms, Z 2 = .3 henry, 

C 3 = .000025 farad, 

J? A = 100 ohms. • 

With a frequency of 60 cycles — whence (0 = 377 — it is 
required to find the pressure necessary to be applied to the 
circuit to send 10 amperes through it. 




Fig. 51. 



The completion of the successive parallelograms in 
Fig. 51, is equivalent to completing the impedance poly- 
gon, and the parts are so marked as to require no explana- 
tion. The solution shows that the equivalent impedance, 
Z=220,.5 ohms, that the equivalent resistance (= actual 
resistance in series), R = 22$ ohms, that the equivalent re- 
actance is condensive and equals 46. 2 ohms, and that eft = 



ALTERNATING-CURRENT CIRCUITS. 87 

1 1.5 5 of lead. Hence the pressure required to send 10 
amperes through the circuit is 

E = 10 x 229.5 — 22 95 v °lts. 
To obtain the same results analytically 

E = 10 V[85 + 4o+ioo] 2 + [(94.2 + 113. 1) — (147.3 + 106. 2)J 2 
E = 2295 volts. 

The voltages at the terminals of the various pieces of ap- 
paratus are : 

E 1 = 10 V85 2 +(94.2 — 147. 3) 2 = 1001 volts, 
E 2 = 10 V40 2 -f- 1 13. i 2 = 1200 " 

^3=IO Vo 2 + I06. 2 2 = I062 " 

E± = 10 Vioo 2 -f o 2 = 1000 " 

E 1 + E 2 J t E z + E 4t = ^6~ 3 " 

which is greater than ^ = 2195 volts, showing that the 
numerical sum of the pressures is greater than the im- 
pressed pressure ; while the vectorial sum of the separate 
pressures is equal to the impressed pressure. 

39. Polygon of Admittances. — If a group of several 
impedances, Z x , Z^ etc., be connected in parallel to a 
common source of harmonic E.M.F. of E volts, their 
equivalent impedance is most easily determined by con- 
sidering their admittances Y lf F 2 , etc. The currents in 
these circuits would be 

I X = EY X , 

I, = EY 2 . 

The total current, supplied by the source, would be the 
vector sum of these currents, due consideration being given 
to their phase relations. Calling this current /, the equation 
I=EY can be written, where Fis the equivalent admit- 



88 



ALTERNATING-CURRENT MACHINES. 



tance of the group. To determine Y, a geometrical addition 
of Y v Y 2 , etc., must be made, the angular relations being 
the same as the phase relations of I v I 2 , etc., respectively. 
The value of the equivalent admittance may therefore be 
represented by the closing side of a polygon, whose other 
sides are represented in magnitude by the several admit- 
tances Y v F 2 , etc., and whose directions are determined 
by the phase angles of the currents I v I 2 , etc., flowing 
through the admittances respectively. 

Fig. 52 is a polygon of admittances showing the method 
of obtaining the equivalent admittance graphically for a 




Fig. 52. 



number of admittances in parallel. The equivalent admit- 
tance may also be determined analytically, since 



Y = vY + b\ 
and hence it follows that the current 



I = EV( gl +g 2 + ...) 2 + {b, + b 2 + ,..) 2 , 

where g v g 2 , . . . and b v b 2 , . . . are the respective con- 
ductances and susceptances of the various pieces of appa- 
ratus. 

The instantaneous value of the current in the main cir- 
cuit is equal to the sum of the instantaneous current values 



ALTERNATING-CURRENT CIRCUITS. 89 

in the branch circuits, but since their maximum values occur 
at different times, the sum of the effective values of current 
in the branches generally exceeds the effective current value 
in the mains. 

As a numerical example on the foregoing, consider the 
same apparatus as was used in the R L c 

preceding example (§ 38), to be arranged r- §$$W&*-i 
in parallel, as in Fig. 53. It is required | v?rB ^ r ^ 
to find the current that will flow through - 
the mains when a 60^ alternating E.M.F. 
of ten volts is impressed upon the L/ WvWV u 

r r Fig. 53- 

circuit. 

7? JT 

Remembering that g = — and that b = — and referring 

Z-j Zj 

to § 38 for the numerical values, the conductances and 
susceptances of the branch circuits are 



* - w - -°° 848 




\ - - S3 -\ - - 
100. r 


•oo53 


4° 

s 2 =— r =.00278 




b = i^i- 
120 2 


.00786 


&= ° 




, I06.2 
3 " "l06.2 2 " 


.00942 


100 
*'- xoo* - - OI 




b 4 = 




Adding algebraically, 








g = .02126 




b = — .00686 




Then Y = V (.02126 V 


5 + 


(- .00686 ) 2 = .025 


?4. 



Hence I = EY = 10 X .0224 = .224 amperes. 



9 o 



ALTERNATING-CURRENT MACHINES. 



The phase of / is given by 

, . b _< — .00686 Q . 

<£ = tan J - = tan * = - 17 53', 

g .02126 

the negative sign indicating that the current leads the 
E.M.F. The admittance of each branch circuit and the 
value and phase of the current therein, may be calculated 
by proceeding in a similar manner. 



40. Impedances in Series and in Parallel. — If a circuit 
have some impedances in series and some in parallel, or in 
any series parallel combination, the equivalent impedance 
can always be found by determining the equivalent impe- 
dances of the several groups, and then combining these 
resulting impedances to get the total equivalent impedance 
sought. To illustrate, a problem will be worked out in 
detail. 

Let it be required to determine the values and phases of 
the currents in the main and in each of the four branch 




Fig. 54- 

circuits, A, B, C and D of the combination shown in 
Fig. 54, when the main terminals are connected to a 200-volt 
25-cycle alternator. 

The constants of the apparatus and the results of the 



ALTERNATING-CURRENT CIRCUITS 



91 



various steps in the calculation are given below in tabulated 
form, and require no explanation. 





A 




B 


C 




D 




R 


50 




20 







150 


ohms 


L 


1 




■5 







6 


henrys 


C 


.00002 







.00001 




.000008 


farads 


(oL 


157.08 




78.54 







942.48 




1 


318.31 




O 


636.62 




795-78 




X 


— 161.23 




78.54 


— 636.62 




146.70 




z 


168.80 




8I.05 


636.62 




209.80 




$ 


- 72° 46' 




75° 43' 


-90° 




44° 3%' 




Y 


•O0593 




.01234 


.001571 




.00477 




g 


.001758 




.00305 







.00339 




b 


— .00566 




.01195 


- -0^1571 




•00335 






Sa+b 


= 


.00481 




Sc+d = - 00 339 




b A + B 


= 


.00629 




b c+D = .00178 




Yab 


= 


.00792 




* CD 


= .00383 




&AB 


= 


52° 36' 




&CD 


= 27° 


42' 




7 

^AB 


= 


126.3 




7 

^CD 


= 261 


.2 




R AB 


= 


76.74 




R CD 


= 231 


•3 




X AB 


= 


100.3 




*CD 


= 121 


•4 






«t 


= 308.0 = 


total equiv. 


resistance 








x , 


= 221.7 = 


total equiv. 


reactance 








z , 


= 379-5 = 


total equiv. 


impedance 








i, 


= .529 = 


current in mains 










*, 


= 35° 45' = 


= phase of 1 


r 








F — 

^AB 


Z AB I, = 66 ' 81 


F = 

^CD 


Z CD T t = - 


[38.17 






r A 


= -396 


*C 


= .217 





.824 



•649 



It is evident that the sum of the potential differences 
across the two groups is greater than the impressed E.M.F., 



92 



ALTERNATING-CURRENT MACHINES. 



and that the arithmetical sum of the currents in the branch 
circuits exceeds the total current flowing in the main cir- 
cuit. The relative magnitudes and the phases of the 



*& 



*3&' \ \ \27°42' 
J LI I 



',44°3 8 > " 5 29 A. 

&9 



Sfei 



PHASE OF I. 



i.V 



\\ 



\\ 



\ 



Fig. 55- 

various currents and E.M.FJs in the different circuits are 
represented in Fig. 55, which also serves as a rough check 
upon the calculations. 



PROBLEMS. 

1. A 60 ~w alternating E.M.F. of 200 volts maximum is impressed 
upon a circuit having 120 ohms resistance, an inductance of 1 henry and 
a capacity of 25 microfarads. Determine the value and phase of the 
current flowing in the circuit. 

2. What are the values of X, Z } Y, b, and g in the preceding problem ? 



PROBLEMS. 93 

3. If a harmonic E.M.F. of 220 volts (effective) is impressed upon a 
circuit, producing a current of 20 amperes lagging 30 , what will be the 
resistance and reactance of the circuit? 

4. Find the instantaneous current value in the circuit of Problem 1, 
if seconds after impressing the harmonic E.M.F. upon it. 

5. What is the resonant frequency of a circuit having 10 microfarads 
capacity and an inductance of .352 henry ? What will be the drop across 
the condenser at resonance, when 10 amperes flow through it? 

6. A condenser of .003 mf. capacity is charged 20 times per second 
to a potential of 1000 volts. What is the mean effective current value 
of the discharge in a circuit having an inductance of 2 millihenrys, if 
the decrement of the oscillations is .2? 

7. Determine the pressure required to send 10 amperes through the 
circuit shown in Fig. 50, if the frequency is 25 ~w per second, the values 
of the resistances, inductances and capacities being the same as in § 38. 
Give also a graphic solution. 

8. In the arrangement shown in Fig. 53, using the same impedances 
as in the preceding problem, what current will traverse the mains, if a 
25-cycle alternating E.M.F. of 10 volts is impressed thereon ? 

9. If the 200- volt 25 ~~ alternator of § 40 were replaced by another of 
the same voltage but of 60^ frequency, what would be the values and 
phases of the currents in the main and in the branch circuits ? 

10. Let the impedance D of the preceding problem be disconnected 
from the circuit. Then determine the values and phases of the currents 
in the main and in the branch circuits A and B. Construct a vector 
diagram showing the voltage and current relations. 



94 ALTERNATING-CURRENT MACHINES. 



CHAPTER V. 

ALTERNATORS. 

41. Alternators. — An alternator is a machine used for 
the conversion of mechanical energy into electrical energy, 
which is delivered as alternating current, either single- 
phase or polyphase. Alternators, like direct-current gen- 
erators, have field-magnets and an armature; but the 
commutator of the direct-current machine is replaced in 
alternators by slip-rings, which deliver alternating current 
to brushes rubbing upon them when the armature rotates, 
or receive direct currents for exciting the field-magnets 
when they rotate. A polyphase alternator produces two 
or more single-phase alternating E.M.F.'s, which in opera- 
tion send currents in circuits which may or may not be in 
electrical connection with each other. The only relation 
between these E.M.F.'s is that of time, that is, they differ 
in phase. These phase differences depend upon the 
relative positions of the armature windings and may be 
anything from o° to 360 , but it is customary to place them 
so as to produce E.M.F.'s differing symmetrically in phase. 
In two-phase or four-phase alternators, the E.M.F.'s are 
90 apart, in three-phase alternators 120 apart, and in six- 
phase alternators 6o° apart. 

As it is the relative motion of the armature and field- 
magnets which is essential in the generation of E.M.F., it 
is quite as common to have the field-magnets of an alternator 



ALTERNATORS. 95 

revolve inside the armature as to have the armature revolve; 
in fact, nearly all large alternators are of the revolving- 
field type, the revolving-armature type being now generally 
restricted to smaller units. The chief advantage of the 
revolving-field type alternator is that it avoids the collection 
of high-tension currents through brushes, since the arma- 
ture connections are fixed, and only low-tension direct 
current need be fed through the rings to the field-coils. 
Other advantages are increased room for armature insula- 
tion, and, in polyphasers, the avoidance of more than two 
slip-rings. Revolving-field alternators have been con- 
structed to generate 25,000 volts, whereas the E.M.F. 
produced in the revolving-armature type is practically 
limited to 5000 volts. In a few instances, notably at 
Niagara Falls, the field-magnets revolve outside the arma- 
ture. 

Besides the two types mentioned, there is the inductor 
type of alternator, which has both its armature and its 
field windings stationary and has an iron rotating member 
termed an inductor. In this type there are neither brushes, 
collector rings, nor moving electrical circuits. 

It is necessary that all but the very smallest alternators 
should be multipolar to fit them to commercial require- 
ments. For alternators must have in general a frequency 
between 25 and 125 cycles per second; the armature 
must be large enough to dissipate the heat generated at 
full load without its temperature rising high enough to 
injure the insulation; and finally, the peripheral speed of 
the armature cannot safely be made to exceed greatly a 
mile a minute. With these restrictions in mind, and 
knowing that a point on the armature must pass under 
two poles for each cycle, it becomes evident that alterna- 



96 ALTERNATING-CURRENT MACHINES. 

tors of anything but the smallest capacity must be multi- 
polar. 

42. Electromotive Force Generated. — In § 13, vol. i., it 
was shown that the pressure generated in an armature is 

E av = ^ 2P$S ~ ID" 8 , 

00 

where p = number of pairs of poles, 

<£ = maxwells of flux per pole, 
V = revolutions per minute, 

and S = number of inductors. 

In an alternating circuit E = k x E av , where \ is the 
form-factor, i.e., the ratio of the effective to the average 
E.M.F. Hence in an alternator yielding a sine wave 
E.M.F., 

E = 2.22 p®S—- io -8 . 
60 

Inasmuch as p — - represents the frequency, /, 
60 

E = 2.22 $Sj io" 8 . 

An alternator armature winding may be either concen- 
trated or distributed. If, considering but a single phase, 
there is but one slot per pole, and all the inductors that are 
intended to be under one pole are laid in one slot, then 
the winding is said to be concentrated, and if the inductors 
are all in series the above formula for E is applicable. 
Nearly all engine-driven alternators have six slots per pole 
although twelve slots per pole are used when the output 
per pole is large and a long armature is undesirable. If 



ALTERNATORS. 97 

now the inductors be not all laid in one slot, but be dis- 
tributed in n more or less closely adjacent slots, the E.M.F. 

generated in the inductors of any one slot will be - of that 

generated in the first case, and the pressures in the differ- 
ent slots will differ slightly in phase from each other, since 
they come under the center of a given pole at different 
times. The phase difference between the E.M.F. gener- 
ated in two conductors which are placed in two successive 
armature slots, depends upon the ratio of the peripheral 
distance between the centers of the slots to the peripheral 
distance between two successive north poles considered as 
360 . This phase difference angle 

_ width slot -f width tooth 
circumference armature ." 



no. pairs poles 



If the inductors of four adjacent slots be in series, and 
if the angle of phase difference between the pressures 
generated in the successive ones be <f>, then letting E lt E 2 , 
E s , and E 4 represent the respective pressures, which are 




Fig. 56. 



supposed to be harmonic, the total pressure, E, generated 
in them is equal to the closing side of the polygon as 
shown in Fig. 56. Obviously E < E x -f- E 2 -f- E s -f E^ 
If the winding had been concentrated, with all the induc- 
tors in one slot, the total pressure generated would have 
been equal to the algebraic sum. 



9 8 



ALTERNATING-CURRENT MACHINES. 



1.00 





99 






















98 






















.97 




















c 


,96 




















c 
o 
O 


95 




















c 


.94 




















-£ 


.93 




















U 


.92 






















91 


















^ 




.90 




F 


ractic 
Oc 


n of 
:upie 


Pole 
dby 


Distar 
Slots 


ce 




N 



1 Slot. 



The ratio of the vector sum to the algebraic sum of 
the pressures generated per pole and per phase is called the 
distribution constant. Not only may the number of slots 
under the pole vary, but they may be spaced so as to 
occupy the whole surface of the armature between succes- 
sive pole centers (the peripheral distance between two 
poles is termed the pole distance), or they may be crowded 

together so as to 
occupy only one- 
half, one-fourth, or 
any other fraction 
of this space. Both 
the number of slots 
and the fractional 
part of the pole dis- 
tance which they 
occupy affect the 
value of the distri- 
bution constant. A 
set of curves, Fig. 
57, has been drawn, 
showing the values of this constant for various conditions. 
Curves are drawn for one slot (concentrated winding), 2, 
3, 4 slots in a group, and many slots (i.e., smooth core 
with wires in close contact on the surface). The ordinates 
are the distribution constants, and the abscissae the frac- 
tional part of the pole distance occupied by the slots. 

The distribution constant, k 2 , must be introduced into 
the formula for the E.M.F. giving 



2 Slots. 



.3 
Fig. 57. 



3 Slots. 

4 Slots. 
Many 
Slots. 



E = 2 k-Ji^pQS— io" 

DO 



or, for sine waves, 



E = 2.22 k 2 ®Sf io" 



ALTERNATORS. 



99 




43. Armature Windings. — There are separate and 
distinct windings on the armature core of a polyphase 
alternator for each phase, and these may each be separately 
connected to an outside circuit through a pair of terminals, 
or they may be connected to- 
gether in the armature accord- 
ing to some scheme whereby 
one terminal will be common 
to two phases. Some simple 
diagrams of the armature 
windings of multipolar alter- 
nators are given in the ac- 
companying figures. Fig. 58 
shows a single-phase concen- 
trated winding, with the wind- 
ing necessary to render it 
two-phase indicated by dotted 
lines. If the two windings be electrically connected where 
they cross at the point P, the machine becomes a star- 
connected four-phase or quarter-phase alternator. 

Three-phase alternators might be provided with six slip- 
rings or terminals, thus supplying three distinct circuits 
with single-phase alternating current, or with four slip-rings 
or terminals, one of which should be connected to a common 
return wire for the three currents. These are uncommon, 
however, since the usual practice is to provide only three 
slip-rings or terminals, each connected wire acting as a 
return path for the currents flowing in the other two. It 
follows, then, that the current in one wire of a three-phase 
system at any instant is equal and opposite to the sum of 
the currents in the other two wires at that instant. This is 
shown in Fig. 59, where the dotted curve, representing the 



/ 4- POLE. \ 

• SINGLE PHASE. \ 

CONCENTRATED. 
ADDITION OF DOTTED WINDING MAKES IT 
TWO PHASE 

Fig. 58. 



:oo 



ALTERNATING-CURRENT MACHINES. 



sum of the two current curves, is exactly equal and opposite 
to the third current curve. 

There are two methods of connecting the armature wind- 
ings of three-phase alternators which are called respectively 
F-and A-connections. In the first, one end of each winding 




Fig. 59- 



is connected to a slip-ring or terminal; the other ends being 
joined together form a neutral connection, which sometimes 




4- POLE. 

3- PHASE. 

Y. 

CONCENTRATED. 



4- POLE. \ 

3- PHASE. 

CONCENTRATED. 



Fig. 6o. Fig. 6i. 

is connected with a fourth slip-ring or terminal adapting the 
alternator for use with a three-phase, four-wire system. In 
armatures having a A-connection, the three windings are con- 
nected together in series to forma closed circuit, each junction 
being connected to a slip-ring or terminal post. A three-phase, 
F-connected, concentrated armature winding is shown in 
Fig. 6o, and the same when A-connected is shown in Fig. 6i. 



ALTERNATORS. 



IOI 



In these winding diagrams the radial lines represent the 
inductors, and the other lines the connecting wires; the 
inductors of different phases being drawn differently for 
clearness. Where but one inductor is shown, in practice 
there would be a number wound into a coil and placed in 
one slot. For simplicity all the inductors of one phase are 
shown in series. Alternator armatures with distributed 
windings can also be represented diagrammatically similar 
to the foregoing, but the diagrams become very complex 
when there are many slots per pole per phase. For sim- 




Fig. 62. 

plicity, a rectified diagram is given, Fig. 62 representing the 
armature winding of a three-phase alternator. There are 
five slots per pole per phase. 

44. Voltage and Current Relations in Two-Phase Systems. 

— A two-phase alternator may be considered as two sepa- 
rate single-phase alternators of the same size, the E.M.F.'s 
of which are maintained at a phase difference of 90 . The 
maintenance of the phase relation might be accomplished by 
mounting the two armatures on the same shaft and then 
placing the coils in the same relative position with the two 



102 



ALTERNATING-CURRENT MACHINES. 



field-magnets displaced from each other by ninety degrees, 
or with the field-magnets in the same relative position and 
the armature coils displaced by ninety degrees. Let these 
two alternators be denoted by i and 2, Fig. 63, and assume 
that the E.M.F. of alternator No. 2 lags 90 behind that 
generated in alternator No. 1. Then the time variations 



El 





Fig. 63. 

of their E.M.F.'s may be graphically represented as shown. 
This condition is also represented by the vectors E l and 
E 2 , lag being clockwise. Let the effective values of the 
E.M.F.'s produced in either armature winding be E volts 
and the effective current value therein be / amperes. Then, 
since the two circuits of a two-phase, four-wire system are 
electrically distinct, the voltage across each is E volts, their 





Fig. 64. 



phase relations being shown by the vectors, and the current 
in each line wire is / amperes. 



ALTERNATORS. 



103 



Now consider these two alternators to be connected as 
shown in Fig. 64, thus forming a two-phase, three-wire 
system. The other conditions remaining unaltered, the 
E.M.F.'s across AB and BC will be the same as before or 
E volts, and the current flowing in A or C will similarly be 
/ amperes. The voltage across AC is due to the E.M.F.'s 
produced in both alternators, and its instantaneous value 
is equal to the algebraic sum of their simultaneous E.M.F.'s. 
The curves showing the time variation of these instan- 
taneous values are annexed, and E AC is seen to lag 45 
behind E AB . These conditions 
may also be represented by 
vectors as in Fig. 65, and there- 
from 



'AC 



— E AB © E BC — \ 




Fig. 65. 



which is therefore the voltage 
across the lines A and C, and 
it lags 45 behind E AB . 

It should be noted that if 
E 2 leads E t by 90 , then E AC 
will lead E AB by 45 ; and fur- 
ther, if the terminals of the receiving apparatus be re- 
versed, the phases of the E.M.F.'s sending current through 
them will be reversed. Assuming load to be applied 
between A and B, and B and C only, and further that 
the circuits are balanced, the current in line C will then 
lag 90 behind the current in A, as shown in Fig. 65. 
Knowing that the instantaneous value of the current in 
line B is equal and opposite to the sum of the instanta- 
neous current values in lines A and C, its value and phase 
may be determined by adding — I A and — I a vectorially 



104 



ALTERNATING-CURRENT MACHINES 




as shown. Thus I B is seen to be equal to V2 / and to lag 
2 2 5 + <f> behind E AB . 

45. Voltage and Current Relations in Three-Phase 
Systems. — Consider a three-phase, F-connected alternator 
to consist of three single-phase genera- 
tors whose E.M.F.'s are maintained at a 
successive phase displacement of 120 
(§ 44), their external connections being 
as shown in Fig. 66. Let the directions 
of the E.M.F.'s in the three armature 
coils as their axes successively pass a given 
fixed point, be positive, and let these con- 
ditions be indicated by the small arrows. 
Then, the phase relations of the armature 
electromotive forces will be represented 
as in Fig. 67, in which E 3 
lags behind E 2 , and E 2 lags 
behind E v The potential 
differences across the various 
line wires may then be de- 
termined by vectorial addi- 
tion and subtraction; for 
example, the E.M.F. across 
AB is equal to the vectorial 
difference of E t and E 2 , since 
they are oppositely directed. 
Taking the momentary posi- 
tive flow as directed towards A, then 



Fig. 66. 




Fig. 67. 



Similarly 



Eab = E 1 e E 2 and leads E 1 by 30 . 



E BC = E 2 e E 3 and lags 90 behind E v 
E C A = £ 3 © ^1 and lags 210 behind E v 



ALTERNATORS. 



105 



Calling the E.M.F. generated in each armature E volts as 
before, the magnitudes of E AB , E BC , and E CA will each be 
V3 E, as may readily be proven by geometry. As the 
current flowing in each line wire is the same as that in each 
armature, it will be / amperes, and if the circuits are 
balanced, i.e. if three loads, each having the same resist- 
ance and the same reactance, are connected respectively 
between A and B, B and C, and C and A, the phases of the 
currents in them will be 120 apart, as shown. Therefore, 
in a three-phase, F-connected system, the voltage between 
any two line wires is V3 E volts, and the current in each 
line is I amperes. 

Now let these three alternators be connected as in Fig. 
68, thus forming a three-phase mesh- or A-connected 





Fig. 68. 

system. The E.M.F. across two line wires is produced by 
one alternator only and is therefore E volts, and if all the 
other conditions remain unchanged, E BC will lag 120 
behind E AB , and E CA will lag 120 behind E BC . Assuming 
the three phases to be equally loaded, and representing 
positive current flow in the coils as their axes successively 
pass a fixed point by the small arrows, the magnitudes of 



io6 



ALTERNATING-CURRENT MACHINES. 



the currents in the lines may be determined vectorially as 
in Fig. 68, where $ is the angle of lag. Hence 

I a = I t © ^ 3 an d lags 30 + (f) behind E v 
Ib = 1 2 © h an d lags 150 + (j) behind E v 
and I c = I 3 e I 2 and leads E t by 90 - <p, 

the magnitude of each being V3 /. Then, to sum up, the 
voltage between any two lines in a balanced three-phase, 
A-connected system is E volts, and the current in each line 
wire is V3 I amperes. 

The power delivered by a three-phase alternator is 
independent of the manner of connection, for in one case 
each leg is supplied with I amperes at V3 E volts, and 
in the other case with V3 I amperes at E volts. 

46. Voltage and Current Relations in Four-Phase Systems. 

— To obtain the current and voltage relations in four- 




E DA E 


4 E AB 


f\ ' 


"/I t 


fc 3K 


\ > !e ^e 
\ 1 * 


» > 


E 2 E BC 



Fig. 69. 



phase systems, consider the four-phase alternator to con- 
sist of four single-phase alternators whose E.M.F.'s are 



ALTERNATORS. 



107 



maintained ninety degrees apart successively. When these 
alternators are star-connected as in Fig. 69, it will become 
evident from an inspection of the vector diagram that the 
voltages between line wires are as follows, the order of the 
subscripts denoting momentary positive direction: 

E A b = E l e E 2 = V 2 E and leads E x by 45 , 
E BC = E 2 e E 3 = V2 E and lags 45 behind E v 
E CD = E 3 e E 4 = V2 E and lags 13 5 behind E v 
E D a = £ 4 O E x = V2 E and lags 225 behind E v 
E AC = E t e E 3 = 2 E and is in phase with E v 



'BD 



E 2 e E4 = 2 E and lags 90 behind E v 



If the circuits are balanced, the current in each line wire is 
the same as that flowing in an armature winding, or / 
amperes. 

When the four single-phase alternators are ring-connected 
as in Fig. 70, the voltage across adjacent_line wires is E 
volts, and across alternate line wires is V2 E volts. The 

current in each line wire is V2 / amperes, 1 — - a 

and the phases of these currents are rep- 
resented in Fig. 71. 

The relations of the voltages and cur- 
rents in the armature windings of a six- 
phase alternator to the voltages across 
the line wires and to the currents therein 
may similarly be determined. 

47. Measurement of Power. — The 

power delivered to the receiving circuits 

of a two-phase, four-wire system can be 

measured by two wattmeters, one con- Fig. 70. 

nected in each phase. The sum of their readings is the 

total power supplied. If the load is balanced, one of the 




io8 



ALTERNATING-CURRENT MACHINES. 



«r ¥ > 



J 2 
Fig. 71. 



wattmeters may be dispensed with, and the total power is 
then double the reading of the other. 

In any two-phase, three- 
wire system the power can 
^be measured by two watt- 
meters connected as in Fig. 
72. The sum of the instru- 
ment readings is the whole 
power. In a two-phase, 
three-wire system, where all 
the load is connected be- 
tween the outside wires and 
the common wire, and none 
between the outside wires themselves, and where the load 
is balanced, then one wattmeter can be used to measure 
the whole power by connecting its current coil in the 
common wire and its pressure-coil between the common 
wire and one outside wire first, then shifting this con- 
nection to the other 
outside wire, as in- 
dicated in Fig. 73. 
The sum of the in- 
strument readings in 
the two positions is 
the whole power. 
A wattmeter made 
with two pressure- 
coils could have one connected each way, and the instru- 
ment would automatically add the readings, giving the 
whole power directly. Or, again, a high non-reactive resis- 
tance could be placed between the two outside wires 
and the pressure-coil of the wattmeter connected between 




Fig. 72. 



ALTERNATORS. 



IO9 



the common wire and the center point of this resistance. 
This requires that the wattmeter be recalibrated with half 
of this high resistance in series with its pressure-coil. 

With the exception of the two-phase systems, the power 
in any balanced polyphase system may be measured by one 
wattmeter whose current coil is placed in one wire, and 
whose pressure-coil is connected between that wire and the 
neutral point. The 






n 




Fig- 73. 



instrument reading 

multiplied by the 

number of phases 

gives the whole power. 

The neutral point 

may be on an extra 

wire, as in a three-phase, four-wire system; or may be 

artificially constructed by connecting the ends of equal 

non-reactive resistances together, and connecting the free 

ends one to each of the phase wires. 

With the exception of the two-phase systems, the power 
in any w-phase, n-wire system, irrespective of balance, may 
be determined by the use of n — 1 wattmeters. The 
current coils are connected, one each, in n — 1 of the wires, 
and the pressure-coils have one of their ends connected to 
the respective phase wires, and their free ends all connected 
to the nth. wire. The algebraic sum of the readings is the 
power in the whole circuit. Depending upon the power 
factor of the circuit, some of the wattmeters will read 
negatively, hence care must be taken that all connections 
are made in the same sense; then those instruments which 
require that their connections be changed, to make them 
deflect properly, are the ones to whose readings a negative 
sign must be affixed. 



no 



ALTERNATING-CURRENT MACHINES. 



Some specific connections for indicating wattmeters in 
three-phase circuits are shown in the following figures. 
Fig. 74 shows the connection of three wattmeters to meas- 
ure the power in an unbalanced three-phase system. All 




££ 



Fig. 74- 

the readings will be in the positive direction, and their 
sum is the total power. If a fourth, or neutral wire be 
present, it should be used, instead of creating an artificial 
neutral, as shown. The magnitude of the equal non- 
reactive resistances, used to secure this neutral point, 
must be so chosen that the resistances of the pressure-coils 
of the wattmeters will be so large, compared thereto, as 
not to disturb the potential of the artificial neutral point. 

Fig. 75 shows the con- 
nection of one wattmeter, 
so as to read one-third 
of the whole power in 
a balanced, three-phase, 
four-wire system. If the 
system be three-wire, a 
neutral point may be 
created as in Fig. 74. 

Another method of measuring power in a balanced three- 
phase system, either A-connected or F-connected, is based 
upon the assumption that both pressures and currents vary 




Fig- 75- 



ALTERNATORS. 



Ill 




Fig. 75. 



harmonically. No neutral point is required, and the con- 
nections are shown in Fig. 76. The free end of the pressure- 
coil is connected first to one of the wires other than that in 
which the current 
coil is connected, 
and then to the other. 
The angular dis- 
placements between 
the current in any 
line wire and the 
E.M.F.'s between 
it and the other line wires are 30 + <f> and 30 — <£, .as 
will become evident from an inspection of Fig. 67 and 
Fig. 68. The readings of the wattmeters are then 

P t = V3 £7 cos (30 + </>) 
and P 2 = V3 EI cos (30 - 0) 

where E is the E.M.F. generated and I is the current flowing 
in each armature coil. The algebraic sum of the read- 
ings is 
p i + P 2 = ^lEI [cos ( 3 o° + <f>) + cos ( 3 o°_- $)] 

= V3 EIHV3 cos<f>— fsin^ + JV^ cos <£ -f- J sin <£] 

= 3 EI cos <f> 

which is the total power delivered. When <f> is greater than 
6o°, Pj becomes negative, hence care is required to avoid 
confusion of signs at low power factors. Both readings 
will be positive if the power factor is greater than .5, but one 
of them will be negative if it is less than this value. 

The algebraic difference of the two wattmeter readings is 
P 2 -P 1 = \/3 £7 [cos (30 - cf>) - cos (30 + <£)] 

= ^s_EI [J V3 cos $ + \ sin <f> - £ V3 cos <£ + J sin cj>] 

= V3 EI sin <f). 



112 



ALTERNATING-CURRENT MACHINES. 



It is more convenient, however, to consider line voltages 
and line currents instead of those in the alternator armature 
windings or in the load of each phase. Therefore, repre- 
senting the E.M.F. between any two line wires by E h and 
the current in each line by_//, then, since either E ( = V3 E 
(F-connection) or I t = V3 I (A-connection), by dividing 
the previous results by V3, there is obtained 

P, + P 2 = V 3 EJ, cos<f>, 
P 2 - P x = EJ t sin 4>. 

In the balanced three-phase system under consideration, 
it is possible to determine the power factor of the similar 
receiving circuits by the use of a single wattmeter connected 
as in Fig. 76. The readings of the wattmeter in the two 
positions are the only observations required. The power 
factor is clearly 

a. ■ -■< r - p. - ' p. 

cos 9 = cos tan 



[* fe& 



p. + p. 

which is derived from the two preceding equations. 

An accurate method for the determination of the power 

in unbalanced three- 
phase systems, avoid- 
ing the necessity of a 
neutral point, involves 
the use of two watt- 
meters connected as 
in Fig. 77. The alge- 
braic sum of the in- 
Fl£ ' 77 ' strument indications 

is the total power supplied. It is possible to obtain 
negative readings, but since the currents lag behind their 
respective E.M.P.'s by different amounts in an unbal- 




ALTERNATORS. 113 

anced system, it cannot be said that when the power 
factor is less than 0.5 one instrument reads negatively, for 
the term power factor here has no definite significance. 
To determine, then, the correct signs of the wattmeter 
readings, the given load may be replaced by a non-inductive 
balanced load of lamps, and if the terminals of the potential 
coil of one instrument must now be reversed to deflect prop- 
erly, it shows that the negative sign must be affixed to its 
reading on the load to be measured. 

48. Saturation. — The electromotive force produced in 
an alternator at no-load is dependent upon the peripheral 
speed of the rotating member and upon the field excitation. 
The relation of the open circuit voltage to the field current 
when the alternator is driven at constant speed may be 
represented by a curve called the no-load saturation curve. 
For a certain 65 k.w. two-phase 2400-volt 60-cycle inductor 
alternator, running at 900 revolutions per minute (air-gap 
of 85 mils), the no-load saturation curve has been experi- 
mentally determined, and is shown in Fig. 78, curve A. It 
indicates, for example, that the field current necessary to 
produce the rated voh^ge on open circuit when the machine 
runs at its proper speed is 4.45 amperes. It is seen that this 
curve is almost straight for small exciting currents. At 
small excitation, the reluctance of the air-gap is very high 
and that of the iron very low, and therefore the former may 
be considered as constituting the entire reluctance of the 
magnetic circuit. Since the reluctivity of air is constant 
regardless of the flux density, at small excitations the flux 
will be proportional to the magnetomotive force, and there- 
fore the open-circuit voltage is proportional to the field 
current, hence the curve is straight. As the field becomes 



H4 



ALTERNATING-CURRENT MACHINES. 



stronger, however, the proportion of the air-gap reluctance 
to the entire reluctance decreases, for the permeability of 
iron decreases with increased flux-density, and therefore 
the E.M.F. increases less rapidly with increased excitation. 





















A^ 




SZD0 

2400 

2100 

I 

g 1800 

6 1500 

1 

JJ200 
900' 

/mo 

/ 300 












/ s 






B^ 














// 


































/ 


// 


















// 
















/ 




















/ 
















































































/ 























4 5 6 

FIELD AMPERES 



Fig. 78. 

The percentage of saturation of an alternator at any 
excitation may be found from its saturation curve by draw- 
ing to it a tangent at the assigned excitation and deter- 
mining its intercept on the axis of ordinates. The ratio of 
this intercept to the ordinate of the curve at the assigned 
excitation, expressed as a percentage, is the percentage of 
saturation. For example, the percentage of saturation of 
the alternator mentioned, when the field current is 4.45 
95o 



amperes, is 



2400 



X 100 = 39. 



ALTERNATORS. 115 

The ratio of a small percentage increment of field excita- 
tion in an alternator to the corresponding percentage incre- 
ment of terminal voltage produced thereby, is called the 
saturation factor. Unless otherwise specified, it refers to 
the excitation existing at normal rated speed and voltage 
and on open circuit. The saturation factor is a criterion 
of the degree of saturation and may be expressed as 



1 — m 

where m is the percentage of saturation. Thus, the satura- 
tion factor of the alternator whose saturation curve is shown 
in Fig. 78, is 

— - = 1.66. 

1 - .396 

The relation of the terminal voltage to the field current 
when the alternator is driven at its rated speed and deliver- 
ing its rated current is given by the full-load saturation curve, 
which is somewhat similar in shape to the no-load satura- 
tion curve. It may be determined experimentally by 
employing variable non-inductive resistances for maintain- 
ing the constant full-load current on each phase, and noting 
the terminal voltage corresponding to various excitations. 
The full-load saturation curve for the 65 k.w. inductor 
alternator at unity power factor is shown as curve B in Fig. 
78. As this curve takes into account all of the diverse 
causes of decrease in terminal voltage resulting from the 
application of a load to the machine, it is important in the 
calculation of regulation. In alternators of large capacity, 
it is a difficult matter to determine the full-load saturation 
curve by test, and consequently other methods are usually 
employed. 



Il6 ALTERNATING-CURRENT MACHINES. 

If the alternator is normally excited to above the knee of 
the saturation curve, it will require a considerable increase 
of field current to maintain the terminal voltage when the 
load is thrown on, while, if normally excited below the knee, 
a slight increase of excitation will suffice. 

49. Regulation. — The regulation of an alternator is the 
ratio of the maximum difference of terminal voltage from 
the rated load value, occurring within the range from open 
circuit to rated load, to the rated load terminal voltage, 
the speed and field current remaining constant. As the 
maximum deviation during this range generally occurs at 
the rated load, it is customary to define regulation as the 
ratio of the rise in terminal voltage, that occurs when full 
load at unity power factor is thrown off, to the terminal 
voltage. Or, expressing it in the form of an equation, 

_> , ^ No-load Voltage — Full-load Voltage 

Regulation = ^ „° — rrr^ • 

5 Full-load Voltage 

An alternator having perfect regulation is one which shows 
no increase in terminal voltage upon opening its load circuit, 
that is, the regulation is zero. 

With small machines, the regulation can be easily deter- 
mined by test, provided artificial loads are available. It is 
simply necessary to plot the no-load and full-load saturation 
curves, and from them the regulation at any load can be 
found. Referring to Fig. 78, for example, the regulation of 
the alternator at full load with unity power factor is 

— — = .146 or 14.6 per cent. With large machines, 

2400 

however, artificial loads are not usually available, and the 

determination of regulation in this case is more difficult and 

less accurate. 



ALTERNATORS, I 17 

The factors affecting the voltage drop in an alternator 
upon the application of load thereto, are, the armature 
resistance, armature reactance, and magnetization or 
demagnetization, occurring especially at low power factors. 
These factors are sometimes grouped together and dealt 
with collectively by the use of a quantity called the syn- 
chronous impedance. It is that impedance, which, if 
connected in series with the outside circuit and to an 
impressed voltage of the same value as the open-circuit 
voltage at the given speed and excitation, would permit a 
current of the same value to flow as does flow. 

The armature resistance drop, seldom exceeding three 
per cent of the terminal voltage, is, for each phase, equal 
to the product of the resistance of the armature winding of 
that phase and the current flowing through it. In calcu- 
lating regulation, the hot resistance (at 75°C.) of the 
windings is always taken. 

The armature conductors of an alternator cut the mag- 
netic flux due to the field current, and this flux may be 
considered as sinusoidally distributed at no-load. An 
E.M.F. will thereby be produced, which will cause a current 
to flow through the armature windings and through the 
load circuit. The armature ampere-turns set up a mag- 
netic flux which is superimposed upon the field flux. The 
magnitude and phase of the terminal electromotive force 
will depend upon this resulting flux, and, if that due to the 
field excitation be constant, then the terminal E.M.F. will 
vary in a manner depending upon the flux due to the arma- 
ture current, which, for brevity, will be called the armature 
flux. The armature self-induction, being proportional to 
the armature flux, varies and depends upon the relative 
positions of the armature and the field and upon the mag- 



n8 



ALTERNATING-CURRENT MACHINES. 



nitude and phase of the currents in the armature windings. 
This variation is shown in Fig. 79, which gives the induc- 
tance corresponding to different angular positions of the 




2 

1 



) 80 100 120 

ELECTRICAL DEGREES 



Fig. 79. 

armature, zero degrees representing coincidence of pole and 
coil group center lines. These curves refer to the 65 k.w. 
two-phase inductor alternator (§ 48), with a current of 9 
amperes flowing through the winding of one phase only. 
The upper curve embodies results taken with the field coil 
open-circuited, and the lower one with the field coil short- 
circuited. Thus, in single-phase alternators, the armature 
flux varies in time and in space. 

Consider a polyphase alternator having a revolving 
armature, a distributed armature winding, a magnetic 
circuit yielding a uniform magnetic reluctance as regards 
the flux due to the current in any armature conductor, and 
a balanced load. The armature flux will be approximately 
sinusoidally distributed in space and stationary as regards 
the field winding, for it would revolve backward as fast as 
the armature revolves forward. The axis of the armature 
flux, when the current through the conductors is in phase 
with the E.M.F., is at right angles to that of the field flux, 
as shown in Fig. 80 by the dotted line sn. When the load 
on each phase is inductive, the axis of the armature flux is 
displaced in the direction of rotation; and when the current 



ALTERNATORS. 



II 9 




Fig. 80. 



supplied to the load leads the E.M.F. of the alternator, the 
axis of the armature flux is displaced in the direction oppo- 
site rotation. These con- 
ditions are represented 
by the dotted lines s'n' 
and s"n n respectively. 
From an inspection of 
the figure, it becomes 
evident, that with a non- 
reactive load the arma- 
ture flux neither assists 
nor opposes the field 
flux; with an inductive 
load, the armature flux has a component which is opposite 
to the field flux; and with a capacity load the armature flux 
has a component which is in the same direction as the 
field flux. The magnetomotive force causing the armature 
flux may then be considered as composed of two compo- 
nents, a transverse component, which is a measure of the 
armature inductance, and the magnetizing or demagnetiz- 
ing component, which acts either with or against the field 
magnetomotive force, depending upon the nature of the 
load. 

Commercial alternators do not have a uniform magnetic 
reluctance, a perfectly distributed winding, nor a sinusoidal 
flux-distribution; and therefore an exact theoretical treat- 
ment of alternator regulation is impossible. 

50. E.M.F. and M.M.F. Methods of Calculating Regu- 
lation. — Two methods of calculating the regulation of 
alternators from the results of other than full-load tests 
have been widely employed, but the results are only approx- 
imate. The first, called the E.M.F. method, generally 



120 ALTERNATING-CURRENT MACHINES. 

gives a larger regulation, and the second, called the M.M.F. 
or A.I.E.E. method, generally gives a smaller regulation 
than what is obtained by test. 

The E.M.F. method may be stated as follows: To 
determine the regulation of an alternator when supplying 
a given current to a receiving circuit of unity power factor, 
add the armature resistance drop to the rated terminal 
voltage, and add the sum vectorially at right angles to 
the armature impedance voltage, that is, the open-circuit 
voltage corresponding to the given short-circuit current 
value. This result minus the rated voltage gives the 
voltage rise at the required load, and dividing this by the 
rated voltage, the regulation at that load will be obtained. 

Consider these factors in detail. Instead of expressing 
the armature resistance drop in terms of the resistance of 
each winding and the current therein, it is desirable in 
practice to express it in terms of the line current and the 
resistance between any two armature terminals. For 
example, take a three-phase alternator and assume it to be 
connected to a balanced load. Representing the line 
voltages and line currents respectively by E and /, and the 
resistance of each armature winding by r, then, in a Y- 
connected alternator, the total copper loss is 3 Pr, and in a 

A-connected machine the total copper loss is 3[ — =) r = Pr. 

If R is the armature resistance between terminals, then 

1 2 

R = 2 r in a F-connected alternator, and R = = - r 

1 1 7 

- +— d 
r 2r 

in a A-connected alternator. Hence the total copper loss, 

-Z R 

whether the machine is Y- or A-connected, is - PR or 3 P — 

2 2 . 

To reduce this result to an equivalent single-phase circuit 



ALTERNATORS. 121 

with the same voltage between line wires and representing 
the same power, P, consider that the rated current per 

P P 

terminal in a single-phaser is — , in a two-phaser is — -, and 

E 2 £ 

p 
in a three-phaser is — ==—. Denoting the equivalent single- 
ts £ 
phase current by I eq , it follows that in a single-phase cir- 
cuit I eq = 7, in a two-phase circuit I eq = 2 7, and in a 
three-phase circuit I pq = V3 7. The equivalent single-phase 
copper loss in a three-phase alternator is J P eq R. Dividing 
by I eq , there results the equivalent single-phase armature 
resistance drop of a three-phase alternator, which is J I eq R- 
This result is also true for two-phase star- or ring-connected 
alternators, as may readily be proven. Thus, the armature 
resistance drop in a polyphase alternator is the product of 
the equivalent single-phase current and half the armature 
resistance as measured between terminals. 

The armature impedance voltage is obtained from the 
short-circuit current and the no-load saturation curves. 
The short-circuit current curve represents the field excita- 
tions required to send various currents through the short- 
circuited armature windings, and may be obtained by direct 
test without requiring large power expenditures. The short- 
circuit current curve for the alternator considered in the 
two preceding articles is shown in Fig. 81. 

As a numerical example, let it be required to determine 
the regulation of this 65 k.w. two-phase 2400-volt alter- 
nator at full load with unity power factor, the armature 
resistance between terminals being 5 ohms at 25 C. 

The rated current per terminal of the alternator is 

13.5 amperes, and the equivalent 



2 E 2 X 2400 

PTlt IS 

E 



p 

single-phase current is — =27 amperes. Half of the 



122 



ALTERNATING-CURRENT MACHINES. 



armature resistance measured between terminals at 75 C. is 
f + 50 X 2.5 X .004 = 3 ohms. Hence the armature IR 
drop 1327X3=81 volts and is in phase with the terminal 
voltage, since the power factor of the load is assumed to be 
unity. The sum of the armature resistance drop and the 



< 
z 

§40 

ID 

H 
£T 

1- 
Z 
Ui 

§20 

oc 
£10 

< 
s 
a. 







































































































0123456 789 

FIELD CURRENT 
Fig. 81. 

terminal voltage is 2481 volts. The excitation required to 
produce the rated current (13.5 amperes) is 2.55 amperes, 
as obtained from Fig. 81. From the no-load saturation 
curve of Fig. 78 is found the impedance voltage correspond- 
ing to this excitation, and is 1550 volts. Adding the 2481 
volts and the 1550 volts at right angles, there results the 
voltage that is considered from the standpoint of this method 
to be actually generated in the alternator, 



V(248i) 2 + (1550) 2 = 2924 volts. 

Hence the regulation at full load with unity power factor is 

2024 — 2400 onr 

y ^ — = .218 or 21.8%. 

2400 

The result for the same conditions obtained from the no- 
load and the full-load saturation curves is 14.6%, thus 
showing that the E.M.F. method gives a poorer regulation 
than is obtained by test. 

Let it be required to calculate the regulation of the same 



ALTERNATORS. 



123 



alternator at f full-load by the E.M.F. method, when the 
power factor of the receiving circuits is 80%. 

The rated f full-load current is 10.125 amperes and the 
equivalent single-phase current is 20.25 amperes, hence the 
armature IR drop is 20.25 X 3 = 60.75 volts. The ter- 
minal E.M.F. can be resolved into two components, one in 
phase with and the other 
at right angles to the cur- 
rent. These components 
are respectively 2400 X .8 
or 1920 volts, and 2400 X 
sin cos -1 . 8 = 2400 X .6 — 
1440 volts. The impedance 
voltage as obtained from 
the curves of Figs. 81 and 
78 is found to be 1220 
volts. The result of add- 
ing these E.M.F.'s in their proper phases is the voltage 
actually generated in the alternator, namely 




Fig. 82. 



v / (i92o+ 6o.75) 2 + (1440 + i22o) 2 = 3320 volts, * 
as shown diagrammatically in Fig. 82. The regulation, 
then, at J full-load and 80% power factor is 
3320 — 2400 ^ 

2400 
The M.M.F. method of calculating alternator regulation 
at unity power factor may be stated as follows : — The 
exciting ampere-turns corresponding to the terminal voltage 
plus the armature resistance drop, and the ampere-turns 
corresponding to the impedance voltage, are combined 
vectorially to obtain the resultant ampere turns, and the 
corresponding internal E.M.F. is obtained from the no-load 
saturation curve. The difference between this E.M.F. 



124 ALTERNATING-CURRENT MACHINES. 

and the rated voltage is divided by the rated voltage to 
obtain the regulation. 

As a numerical example, let it be required to calculate 
the regulation of the same alternator at full load with unity 
power factor by the M.M.F. method. 

The field current corresponding to 2400 + 81 volts is 
4.7 amperes, as obtained from the no-load saturation curve 
of Fig. 78. The field current corresponding to the impe- 
dance voltage (1550 volts) is found from the same curve and 
is 2.55 amperes. This value can also be obtained directly 
from Fig. 81; it corresponds to the rated current (13.5 
amperes). Then, adding 4.7 amperes a nd 2.55 amperes 

at right angles, there results V4.7 2 + 2. 5 5 2 or 5.35 amperes. 
The no-load voltage corresponding to this excitation is 
2620 volts Therefore the regulation is 

2620 — 2400 ~ 

— = .0916 or 9.16%. 

2400 

This result is much smaller than that obtained by test 
(14.6%), that is, the M.M.F. method for these conditions 
gives a better regulation than it should. The mean value 
of the results obtained by the E.M.F. and M.M.F. methods 
is 15.5% and agrees fairly well with the experimental result; 
but this is not always true. 

51. Regulation for Constant Potential. — Alternators 
feeding light circuits must be closely regulated to give 
satisfactory service. The pressure can be maintained 
constant in a circuit by a series boosting transformer, but 
it is generally considered better to regulate the alternator 
by suitable alteration of the field strength. 

The simplest method of regulating the potential is to 
have a hand-operated rheostat in the field circuit of the 
alternator, when the latter is to be excited from a com- 



ALTERNATORS 



125 



mon source of direct current, or in the field circuit of the 
exciter, if the alternator is provided with one. The 
latter method is generally employed in large machines, 
since the exciter field current is small, while the alternator 
field current may be of considerable magnitude, and would 
give a large PR loss if passed through a rheostat. 

A second method of regulation employs a composite 
winding analogous to the compound windings of direct- 




Fig. 83. 
current generators. This consists of a set of coils, one 
on each pole. These are connected in series, and carry 
a portion of the armature current which has been rectified. 
The rectifier consists of a commutator, having as many 
segments as there are field poles. The alternate segments 
are connected together, forming two groups. The groups 
are connected respectively with the two ends of a resist- 
ance forming part of the armature circuit. Brushes, 
bearing upon the commutator, connect with the terminals 
of the composite winding. The magnetomotive force of 
the composite winding is used for regulation only, the 
main excitation being supplied by an ordinary separately 
excited field winding. The rectified current in the com- 
posite coils is a pulsating unidirectional current that 
increases the magnetizing force in the fields as the cur- 
rent in the armature increases. The rate of increase is 



126 ALTERNATING-CURRENT MACHINES. 

determined by the resistance of a shunt placed across the 
brushes. By increasing the resistance of this shunt, the 
amount of compounding can be increased. With such 
an arrangement an alternator can be over-compounded to 
compensate for any percentage of potential drop in the 
distributing lines. The method here outlined is used by 
the General Electric Company in their single-phase 
stationary field alternators. The connections are shown 
in Fig. 83. 

A third method of regulation is employed by the West- 
inghouse Company on their revolving armature alter- 
nators, one of which, a 75 k.w., 6o~, single-phase machine, 
is shown in Fig. 84. A composite winding is employed, 
and the compensating coils are excited by current from 
a series transformer placed on the spokes of the armature 
spider. The primary of this transformer consists of but 
a few turns, and the whole armature current is conducted 
through it before reaching the collector rings. The sec- 
ondary of this transformer is suitably connected to a 
simple commutator on the extreme end of the shaft. 
Upon this rest the brushes which are attached to the ends 
of the compensating coil. This commutator is subjected 
to only moderate currents and low voltages. The current 
in the secondary of the transformer, and hence that in 
the compensating coil, is proportional to the main armature 
current. The machine is wound for the maximum desir- 
able over-compounding, and any less compensation can be 
secured by slightly shifting the commutator brushes. 
For there are only as many segments as poles ; and if the 
brushes span the insulation just when the wave of current 
in the transformer secondary is passing through zero, 
then the pulsating direct current in the compounding coil 



ALTERNATORS. 



12/ 



is equal to the effective value of the alternating current ; 
but if the brushes are at some other position, the current 
will flow in the field coil in one direction for a portion of 
the half cycle, and in the other direction for the remaining 




Fig. 84. 



portion. A differential action, therefore, ensues, and the 
effective value of the compensating current is less than it 
was before. 

In order to produce a constant potential on circuits 
having a variable inductance as well as a variable resist- 



128 



ALTERNATING-CURRENT MACHINES. 



ance, the General Electric Co. has designed its compensated 
revolving field generators, which are constructed for two- 
or three-phase circuits. The machine, Fig. 85, is of the 




Fig. 85. 

revolving field type, the field being wound with but one 
simple set of coils. On the same shaft as the field, and 
close beside it, is the armature of the exciter, as shown 
in Fig. 86. The outer casting contains the alternator 
armature windings, and close beside them the field of the 
exciter. This latter has as many poles as has the field of 
the alternator. Alternator and exciter, therefore, operate 
in a synchronous relation. The armature of the exciter is 
fitted with a regular commutator, which delivers direct 
current both to the exciter field and, through two slip- 



ALTERNATORS. 



2 9 



rings, to the alternator field. On the end of the shaft, 
outside of the bearings, is a set of slip-rings, four for a 
quarter-phaser, three for a three-phaser, through which 
the exciter armature receives alternating current from one 
or several series transformers inserted in the mains which 
lead from the alternator. This alternating current is 
passed through the exciter armature in such a manner as 
to cause an armature reaction, as described in § 49, that 
increases the magnetic flux. This raises the exciter vol- 
tage and hence increases the main field current. The 




Fig. 86. 

reactive magnetization produced in the exciter field is 
proportional to the magnitude and phase of the alternating 
current in the exciter armature. The reactive demag- 
netization of the alternator field is proportional to the 
magnitude and phase of the current in the alternator 
armature. And these currents have the fixed relations 
of current strength and phase, which are determined by the 
series transformers. Hence the exciter voltage varies so 
as to compensate for any drop in the terminal voltage. 
Neither the commutator nor any of the slip-rings carry 
pressures of over 75 volts. The amount of over-corn- 



130 



ALTERNATING-CURRENT MACHINES. 



pounding is determined by the ratio in the series trans- 
formers. The normal voltage of the alternator may be 
regulated by a small rheostat in the field circuit of the 
exciter. The various connections of this type of com- 
pensated alternator are shown in Fig. 87. 

The regulation of voltages by means of composite wind- 




Fig. 87. 



ings finds application on alternators up to about 250 k.w. 
output. The Tirrill Regulator, for use with large or small 
generators, is made by the General Electric Company, and 
shown in Fig. 88. This device operates by rapidly opening 
and closing a shunt circuit connected across the exciter 
field rheostat, the operation being accomplished by means 
of a differentially wound relay, which is connected to the 
exciter bus-bars. There are two control magnets, one for 
direct current and the other for alternating current. The 
current for the first is taken from the exciter bus-bars, and 
the current for the latter is taken from a potential trans- 
former connected in the circuit to be regulated. Upon 
the same spool as this potential winding is an adjustable 
compensating winding which is connected to the secondary 



ALTERNATORS. 



131 



of a current transformer inserted in the principal lighting 
circuit. The cores of these control magnets are attached 
to pivoted levers provided with contacts at their other ends. 




Fig. 88. 



When a load is thrown on the alternator, the voltage will 
tend to drop and the alternating-current magnet will 
weaken, thus causing the main contacts to close. This 



132 ALTERNATING-CURRENT MACHINES. 

also causes the relay contacts to close and short-circuit the 
exciter field rheostat, thereby increasing the potential 



A***V COK/7ACTS 
PI\SOT 



etcrrfi 
Rhc-os r/tr. 



AOJirsr/tate 




Fig. 89. 



supplied to the alternator field. The general scheme and 
connections of this regulator for a single generator and 
exciter are shown in Fig. 89. 



ALTERNATORS. 133 

52. Efficiency. — The following is abstracted from the 
Report of the Committee on Standardization of the Ameri- 
can Institute of Electrical Engineers. Only those por- 
tions are given which bear upon the efficiency of alternators. 
They will, however, apply equally well to synchronous 
motors. 

The "efficiency" of an apparatus is the ratio of its net 
power output to its gross power input. 

Electric power should be measured at the terminals of 
the apparatus. 

In determining the efficiency of alternating-current 
apparatus, the electric power should be measured when 
the current is in phase with the E.M.F. unless otherwise 
specified, except when a definite phase difference is in- 
herent in the apparatus, as in induction motors, etc. 

Where a machine has auxiliary apparatus, such as an 
exciter, the power lost in the auxiliary apparatus should 
not be charged to the machine, but to the plant consisting 
of the machine and auxiliary apparatus taken together. 
The plant efficiency in such cases should be distinguished 
from the machine efficiency. 

The efficiency may be determined by measuring all the 
losses individually, and adding their sum to the output to 
derive the input, or subtracting their sum from the input 
to derive the output. All losses should be measured at, 
or reduced to, the temperature assumed in continuous 
operation, or in operation under conditions specified. 

In synchronous machines the output or input should be 
measured with the current in phase with the terminal 
E.M.F. except when otherwise expressly specified. 

Owing to the uncertainty necessarily involved in the 
approximation of load losses, it is preferable, whenever 



134 ALTERNATING-CURRENT MACHINES. 

possible, to determine the efficiency of synchronous ma- 
chines by input and output tests. 

The losses in synchronous machines are : 

a. Bearing friction and windage. 

b. Molecular magnetic friction and eddy currents in 
iron, copper, and other metallic parts. These losses should 
be determined at open circuit of the machine at the rated 
speed and at the rated voltage, + IR in a synchronous 
generator, — IR in a synchronous motor, where I = cur- 
rent in armature, R = armature resistance. It is undesir- 
able to compute these losses from observations made at 
other speeds or voltages. 

These losses may be determined by either driving the 
machine by a motor, or by running it as a synchronous 
motor, and adjusting its fields so as to get minimum cur- 
rent input, and measuring the input by wattmeter. The 
former is the preferable method, and in polyphase ma- 
chines the latter method is liable to give erroneous results 
in consequence of unequal distribution of currents in the 
different circuits caused by inequalities of the impedance 
of connecting leads, etc. 

c. Armature-resistance loss, which may be expressed 
by/ I 2 R ; where R = resistance of one armature circuit 
or branch, / = the current in such armature circuit or 
branch, and / = the number of armature circuits or 
branches. 

d. Load losses. While these losses cannot well be 
determined individually, they may be considerable, and, 
therefore, their joint influence should be determined by 
observation. This can be done by operating the machine 
on short circuit and at full-load current, that is, by deter- 
mining what may be called the "short-circuit core loss." 



ALTERNATORS. 



135 



With the low field intensity and great lag of current 
existing in this case, the load losses are usually greatly 
exaggerated. 

One-third of the short-circuit core loss may, as an 
approximation, and in the absence of more accurate infor- 
mation, be assumed as the load loss. 

e. Collector-ring friction and contact resistance. These 
are generally negligible, except in machines of extremely 
low voltage. 

/. Field excitation. In separately excited machines, 
the PR of the field coils 
proper should be used. 
In self-exciting machines, 
however, the loss in the 
field rheostat should be 
included. 

The efficiency curve of 
an alternator may be 
plotted when the losses at £ 
different loads have been z ° 
determined. The efficiency 2 40 
curve of a ^000 k. w. 
11,000 volt alternator, 
and that of a 1000 k. w. 20 
500 volt alternator are 10 
shown in Fig. 90. 

llo 40 GO 80 100 120 

PERCENT FULL-LOAD 

53. Rating. — Alterna- 
tors are rated by their 

electrical output, either in kilowatts or in kilovolt-amperes. 
By rating is meant the power that the machine can deliver 
to the load without an excessive rise in temperature. This 



O 70 

z 

Ul 

















,/ 


<; 


5000 












1000 V 


..w. 








/ 












/ 

















































































































136 ALTERNATING-CURRENT MACHINES. 

temperature rise is due to the losses in the alternator; these 
include the practically constant iron losses and the copper 
losses, variable with load. Under a fixed current output 
the temperature of the armature will rise until the rate of 
escape of heat from it is equal to the rate of its develop- 
ment. This ultimate temperature should not exceed 8o° C. 
in any case. Under a given voltage the current output is 
limited by the rise of temperature and the power output 
is further limited by the power factor of the load circuit. 
Hence the power supplied by an alternator to a reactive 
load is less than that supplied to a non-reactive load for 
the same temperature rise in the machine. It is advisable 
to rate alternators in kilovolt-amperes and to specify the 
power factor on which this rating is based. Thus, a 6600 
volt alternator whose rated current is 500 amperes, called 
a 3300 kilovolt-ampere alternator, could deliver 3300 kilo- 
watts to a non-reactive load, but, for the same temperature 
rise it could only deliver 2640 kilowatts to a load of 80% 
power factor. 

An alternator should be able to carry a 25% overload for 
two hours without serious injury because of heating, elec- 
trical or mechanical stresses, and with an additional tem- 
perature rise not exceeding 15 C. above that specified for 
rated load, the overload being applied after the machine 
has acquired the temperature corresponding to continuous 
operation at rated load. 

54. Inductor Alternators. — Generators in which both 
armature and field coils are stationary are called inductor 
alternators. Fig. 91 shows the principle of operation of 
these machines. A moving member, carrying no wire, 
has pairs of soft iron projections, which are called indue- 



ALTERNATORS. 



137 



tors. These projections are magnetized by the current 
flowing in the annular field coil as shown in figure. The 
surrounding frame has internal projections corresponding 



/ARMATURE COILS 




/ 



FIELD COIL 



\ 



fr 



Fig. 91. 

to the inductors in number and size. These latter projec- 
tions constitute the cores of armature coils. When the 
faces of the inductors are directly opposite to the faces of 
the armature poles, the magnetic reluctance is a minimum, 
and the flux through the armature coil accordingly a maxi- 
mum. For the opposite reason, when the inductors are 
in an intermediate position the flux linked with the arma- 
ture coils is a minimum. As the inductors revolve, the 
linked flux changes from a maximum to a minimum, but it 
does not change in sign. 

Absence of moving wire and the consequent liability to 
chafing of insulation, absence of collecting devices and 
their attendant brush friction, and increased facilities for 
insulation are claimed as advantages for this type of ma- 
chine. By suitably disposing of the coils, inductor alter- 
nators may be wound for single- or polyphase currents. 

The Stanley Electric Manufacturing Company manu- 
factured two-phase inductor alternators. A view of one of 
their machines is given in Fig. 92, with the frame separated 
for inspection of the windings. In this picture the field 



138 



ALTERNATING-CURRENT MACHINES. 



coil is hanging loosely between the pairs of inductors. The 
theoretical operation of this machine is essentially that 




Fig. 92. 

described above. All iron parts, both stationary and 

revolving, that are subjected to pulsations of magnetic 

flux, are made up of 

laminated iron. The 

large field coil is 

wound on a copper 

spool. Ordinarily 

when the field circuit 

of a large generator 

is broken, theE.M.F. 

of self-induction may 

rise to so high a 

value as to pierce the Fig. 93 

insulation. With this construction the copper spool acts as 

a short circuit around the decaying flux, and prevents high 




ALTERNATORS. 



139 




E.M.F.'s of self-induction. Figs. 93 and 94 show details 
of construction of larger machines of this type. 

55. Revolving Field Alternators. — In this type of al- 
ternator, the armature windings are placed on the inside 



140 



ALTERNATING-CURRENT MACHINES, 



of the surrounding frame, and the field poles project radi- 
ally from the rotating member. As was stated before this 




Fig. 95- 



type of construction is to be recommended in the case of 
large machines which are required to give either high 
voltages or large currents. With the same peripheral 



ALTERNATORS. 



141 



velocity, there is more space for the armature coils; the 
coils can be better ventilated, air being forced through 
ducts by the rotating field; stationary coils can be more 
perfectly insulated than moving ones; and the only cur- 
rents to be collected by brushes and collector rings are 
those necessary to excite the fields. 

Fig. 95 shows a General Electric 750 K. w. revolving 
field generator. The two collector rings for the field cur- 
rent are shown, and in Fig. 96 the edgewise method of 




Fig. 96. 

winding the field coils is shown. The collector rings are 
of cast iron and the brushes are of carbon. Fig. 97 shows 
the details of construction of a 5000 k.w. three-phase 
6600-volt machine of this type as constructed for the 
Metropolitan Street Railway Co. of New York. This 
machine has 40 poles, runs at 75 r.p.m. at a peripheral 
velocity of 3900 feet per minute. This gives a frequency 
of 25. The air gap varies from five-sixteenths at the 
pole center to eleven-sixteenths at the tips. The short- 
circuit current at full-load excitation is less than 800 am- 



142 



ALTERNATING-CURRENT MACHINES. 



peres per leg. The rated full-load current is slightly over 
300 amperes. 

Fig. 98 shows the method of assembling the armature 




SCALE ¥4 INCH EQUJ.'LS ONE FOOT 



Fig. 97. 



coils in the slots of the stationary core. In this machine 
there is a three-phase winding distributed so as to utilize 



ALTERNATORS. 



143 




Fig. 



144 



ALTERNATING-CURRENT MACHINES. 




Fig. 99- 



two slots per pole per phase. Fig. 99 shows the construction 
of a rotating field which consists of a steel rim mounted 
upon a cast-iron spider. Into dovetailed slots in the rim are 
fitted laminated plates with staggered joints. These plates 



ALTERNATORS. 145 

are bolted together. The laminations are supplied at 
intervals with ventilating ducts. The coils are kept in 
place by retaining wedges of non-magnetic material. 

56. Self-Exciting Alternators. — An alternator of a dif- 
ferent type from those previously considered is the Alex- 
anderson self-exciting alternator which is manufactured 
by the General Electric Company. The armature and 
field windings differ in no respect from the usual type used 
in alternating-current generators. The field current is 




Fig. 100. 

derived from an auxiliary winding placed in the same slots 
as the main armature winding, and is rectified by means of 
a segmental commutator with one active segment per pole. 
The revolving field of a 100 k.w. three-phase self-exciting 
alternator with its commutator is shown in Fig. 100. Alter- 
nate segments of this commutator are connected to two 
steel rings surrounding the segments, and these rings are 
connected to the field winding. A terminal of each aux- 
iliary winding is connected to one of three brushes bearing 
on this commutator, and their other terminals are attached 
to a three-phase rheostat, as shown in Fig. 101. 



146 



ALTERNATING-CURRENT MACHINES. 



Automatic compounding is effected by series trans- 
formers connected as indicated in the figure. The amount 
of boosting in the field circuit will depend upon the values 




Fig. 101. 



of the currents in the secondaries of the series transformers 
as well as upon the power factor of the load. 



PROBLEMS. 

1. The armature of a 25 cycle, eight pole single-phase alternator has 
three slots per pole with ten conductors in each slot, the slots occupying 
one-half of the pole distance. If the flux from each pole is 1,200,000 
maxwells, what will be the effective E.M.F. generated in the armature, 
assuming this E.M.F. to be sinusoidal? 

2. Determine the voltage across the outside terminals of a two-phase 
three-wire 100 volt alternator, when the windings on its armature are 
set 85 apart instead of 90 apart. 

3. The E.M.F. generated in each armature winding of a three-phase 
alternator is 125 volts, and the current in each is 5 amperes when the 
alternator is connected to a certain load. Determine the voltages be- 
tween the lines and the current flowing in each line wire, when the 
machine is F-connected and when it is A-connected. 

4. Find the magnitudes and phases of the various voltages across the 



PROBLEMS. 



147 



line wires of a four-phase star-connected system, the voltage generated 
in each armature winding being 75 volts. 

5. A three-phase alternator is connected to a balanced non-reactive 
receiving circuit. It is required to determine the magnitude of the power 
supplied by the alternator, when the voltage across the line wires is 
150 volts and the current in each line is 20 amperes. 

6. A three-phase alternator is connected to a balanced inductive load 
and the power is measured according to the method of Fig. 76. What 
is the power factor of the receiving circuits if the observed indications 
on the wattmeter are 2,200 and 2,900 watts respectively? 



8000 

7000 

£6000 

O 

> 5000 
ixl 

tc 

h-4000 
<. 

<3000 

2000 

1000 

n 










































































































































































































ui 

1 <r 
as cc 

Hi => 

o: a 

11 



100 



20 



40 60 



10 100 120 140 160 180 200 
FIELD AMPERES 



Fig. 102. 



7. In the alternator of Fig. 78, determine the saturation factor when 
the exciting current is six amperes. 

8. Calculate the regulation of the alternator of the preceding problem 
when the field excitation is 7 amperes, and when the power factor of the 
load is unity. 

9. The no-load saturation and the armature short-circuit current 
curves of a 3500 k. w. three-phase 6,600 volt revolving field alternator 
are shown in Fig. 102. Calculate the regulation at full load with unity 



148 ALTERNATING-CURRENT MACHINES. 

power factor by the E.M.F. method. The armature resistance between 
terminals is .093 ohms at 25 C. 

10. Determine the regulation of the alternator of the preceding 
problem on an inductive load of 80% power factor, by the E.M.F. 
method. 

11. If the load of the preceding problem were replaced by a capacity 
load of the came power factor, what would be the regulation at full load, 
as calculated by the E.M.F. method. 

12. Determine the regulation of the alternator of Fig. 102 for each of 
the conditions of the three preceding problems, applying the M.M.F, 
method. 



THE TRANSFORMER. 149 



CHAPTER VI. 

THE TRANSFORMER. 

57. Definitions. — The alternating-current transformer 
consists of one magnetic circuit interlinked with two elec- 
tric circuits, of which one, the primary, receives electrical 
energy, and the other, the secondary, delivers electrical 
energy. If the electric circuits surround the magnetic 
circuit, as in Fig. 103, the transformer is said to be of the 
core type. If the re- 
verse is true, as in 
Fig. 104, the trans- 
former is of the shell 
type. The practical 
utility of the trans- 
former lies in the fact 
that, when suitably 
designed, its primary 
can take electric energy 
at one potential, and 
its secondary deliver 

the same energy at 1S ' I03 ' 

some other potential; the ratio of the current in the primary 
to that in the secondary being approximately inversely 
as the ratio of the pressure on the primary to that on 
the secondary. 

The ratio 0) transformation of a transformer is repre- 




150 ALTERNATING-CURRENT MACHINES. 

sented by z, and is the ratio of the number of turns in the 
secondary coils to the number of turns in the primary coil. 
This would also be the ratio of the secondary voltage to 




Fig. 104. 

the primary voltage if there were no losses in the trans- 
former. A transformer in which this ratio is greater than 
unity is called a "step-up" transformer, since it delivers 
electrical energy at a higher pressure than that at which 
it is received. When the ratio is less than unity it is 
called a " step-down" transformer. Step-up transformers 
find their chief use in generating plants, where because of 
the practical limitations of alternators, the alternating cur- 
rent generated is not of as high a potential as is demanded 
for economical transmission. Step-down transformers find 
their greatest use at or near the points of consumption of 
energy, where the pressure is reduced to a degree suitable 
for the service it must perform. The conventional repre- 
sentation of a transformer is given in Fig. 105. In general, 
little or no effort is made to indicate the ratio of trans- 
formation by the relative number of angles or loops shown, 



THE TRANSFORMER. 151 

though the low-tension side is sometimes distinguished 
from the high-tension side by this means. 

When using the same or part of the same electric cir- 
cuit for both primary and secondary, the device is called 
an auto-transformer. These are sometimes used in the 




¥ 



3- 



Fig. 105. Fig. 106. 

starting devices for induction motors, and sometimes 
connected in series in an alternating-current circuit, and 
arranged to vary the E.M.F. in that circuit. Fig. 106 is 
the conventional representation of an auto-transformer. 

58. The Ideal Transformer. — The term ideal transformer 
may be applied to one possessing neither hysteresis and 
eddy current losses in the core nor ohmic resistance in the 
windings, and all the flux set up by one coil links with the 
other coil also. Actual transformers, however, do not 
satisfy these conditions, yet their behavior approximates 
closely to that of an ideal transformer. 

When the secondary coil of a transformer is open-circuited 
it is perfectly idle, having no influence on the rest of the 
apparatus, and the primary becomes then merely a choke 
coil or reactor. The reactance of a commercial trans- 
former is very large and its resistance very small, con- 
sequently the impedance is high and almost wholly reactive. 
In the ideal transformer the current that will flow in the 
primary when the secondary is open-circuited is very small 
and lags 90 behind the E.M.F. which sends it. This 
current is called the exciting current, and will be sinusoidal 
in the ideal transformer when the impressed electromotive 



152 ALTERNATING-CURRENT MACHINES. 

force is sinusoidal. A flux will be set up in the iron of the 
transformer, which is sinusoidal and in phase with the 
exciting current. This flux induces a sinusoidal E.M.F. 
in the primary coil which is 90 behind the flux in phase 
because the induced E.M.F. is greatest when the time rate 
of flux change is greatest, and this flux change is greatest 
when passing through the zero value. This induced 
electromotive force is 90 behind the flux, which in turn is 
90 behind the impressed E.M.F.) therefore the induced 
E.M.F. is 180 behind the impressed electromotive force, 
or is a counter E.M.F. In the ideal transformer under 
consideration, the counter pressure is exactly equal to the 
E.M.F. impressed upon the primary. The phase relations 
of the pressures, the exciting current, and the flux in an 
ideal transformer are shown in Fig. 107. It should be 
noted that the exciting current, being 
at right angles to the pressure for this 
transformer, does not represent a loss 
in power, for the energy is alternately 
SE received from and supplied to the 

• ux circuit. This may be shown graphi- 

cally as in Fig. 108, where the lobes of 
negative and positive power are equal. 
When the secondary winding of an 
ideal transformer is closed through an 

Fig. 107. & 

outside impedance, the variations in 
the flux, which is linked with the secondary as well as 
the primary, produce in the secondary an E.M.F. r times 
as great as the counter E.M.F. in the primary, since there 
are r times as many turns in the secondary coil as there 
are in the primary, or 

E s = tE p ; 



THE TRANSFORMER. 



153 



and a current I s will flow through the external circuit. The 
ampere-turns of the secondary, nj s , will be opposed to the 
ampere-turns of the primary, and will thus tend to demag- 
netize the core. This tendency is opposed by a readjust- 
ment of the conditions in the primary circuit. Any demag- 
netization tends to lessen the counter E.M.F. in the primary 
coil, which immediately allows more current to flow in the 
primary, and thus restores the magnetization to a value but 
slightly less than the value on open-circuited secondary. 
Thus the core flux remains practically constant whether 




Fig. 108. 



the secondary be loaded or not, the ampere turns of the 
secondary being opposed by a but slightly greater number 
of ampere turns in the primary. So 



and 



n s I s = n p I p , very nearly, 
7 = Hip t = I j 



The counter E.M.F. in the primary of a transformer 
accommodates itself to variations of load on the secondary 



154 



ALTERNATING-CURRENT MACHINES. 



in a manner similar to the variation of the counter E.M.F. 
of a shunt wound motor under varying mechanical loads. 

The vector diagram of the ideal transformer, when the 
secondary is closed through a circuit having a reactance 
X 2 , and a resistance R 2 , is shown in Fig. 109. It represents 
a step-down transformer where z = \. The secondary 

current I 5 lags behind E s by the angle <j> = tan -1 — - 2 . The 

^2 
primary ampere-turns are composed of two components — ■ 
one necessary to balance the secondary ampere-turns and 
the other necessary to magnetize the core, as shown. 

When the transformer delivers little power, the magne- 
tizing component of the magneto- 
motive force is comparable in 
magnitude with the active com- 
ponent, consequently the secon- 
dary current lags behind the 
primary current by less than 180 . 
When the transformer is fully 
loaded, however, the magnetizing 
current is comparatively small 
and therefore the directions of 
I p and 7, are very nearly oppo- 
site. This is also true of com- 
which the exciting current is less 




Fig. 109. 



mercial transformers, in 
than 90 behind E p . 



59. Core Flux. — The relation between the magnetic 
flux in a transformer core and the primary impressed 
E.M.F. can be determined by considering that the flux 
varies harmonically, and that its maximum value is & m ; 
then the flux at any time, /, is 3> m cos cot, and the counter 



THE TRANSFORMER. 



155 



E.M.F., which is equal and opposed to the impressed 
primary pressure E p , may be written (§ 13, vol. i.) 

p_ d (& m cos a)t) 



EJ = 



io c 



dt 



Ep = 10 8 n p a>® m sin cut, 



J? 


= 


10 8 n p oj<P m , 






*m 


= 


io 8 E pm io 8 V 2 
n p oj n p co 


^ 



and since 3> m and w are constant 
from which 

and 

This equation is used in designing transformers and 
choke coils. The values of ® m for 60 cycle transformers 
of different capacities, as determined by experiment and 
use, are shown in the curve, Fig. no. It is usual in such 

" 1.6 
1.4 

I" 

a 

<0 

2 .8 

c 

_3 .6 
Ll 

~£ «4 

7 I I I I 

2 4 6 8 10 12 14 16 18 20 22 24 28 

Capacity in Kilowatts 
Fig. no. 

designs to assume a maximum flux density, (B OT . Trans- 
former cores are worked at low flux densities, and, while 
the value assumed differs considerably with the various 
manufacturers, it is safe to say that for 25 cycles <B W varies 





















































































































Lighting Transformers 
at 60^ 
























/ 


























1 



























156 ALTERNATING-CURRENT MACHINES. 

between 6 and 8 kilogausses; for 60 cycles between 5 and 6 
kilogausses; and for 125 cycles between 3 and 4 kilogausses. 
The necessary cross-section, A, of iron in square centimeters 
is found from the relation 

® m = ®> m A. 

60. Transformer Losses. — The transformer as thus far 
discussed would have 100 % efficiency, no power whatever 
being consumed in the apparatus. The efficiencies of 
loaded commercial transformers are very high, being gen- 
erally above 95 % and frequently above 98 %. The losses 
in the apparatus are due to the resistance of the electric 
circuits, hysteresis, and eddy currents. These losses may 
be divided into core losses and copper losses, according as to 
whether they occur in the iron or the wire of the trans- 
former. 

61. Core Losses. — (a) Eddy current loss. If the core 
of a transformer were made of solid iron, strong eddy cur- 
rents would be induced in it. These currents would not 
only cause excessive heating of the core, but would tend 
to demagnetize it, and would require excessive currents to 
flow in the primary winding in order to set up sufficient 
counter E.M.F. 

To a great extent these troubles are prevented by mak- 
ing the core of laminated iron, the laminae being trans- 
verse to the direction of flow of the eddy currents but 
longitudinal with the magnetic flux. Each lamina is more 
or less thoroughly insulated from its neighbors by the 
natural oxide on the surface or by Japan lacquer. The 
eddy current loss is practically independent of the load. 

An empirical formula for the calculation of the watts 



THE TRANSFORMER. 



157 



lost in the transformer core due to eddy currents, based 
upon the assumption that the laminations are perfectly 
insulated from one another, is 



*2 72rU 2 
in j 



P e = kv) 2 P 
where 

k = a constant depending upon the resistivity of 

the iron, 
v = volume of iron in cm. 3 , 
I = thickness of one lamina in cm., 
/ = frequency, 
and 

(& m = maximum flux density (3> m per cm. 2 ). 

In practice k has a value of about 1.6 X io _n . 

The values of P e in watts per cubic inch and per pound 
in terms of flux density for 25 cycle and 60 cycle transformers 
may be taken directly from the curves of Fig. in. These 

1.0 



0- 

(o .4 



























_ 




ED 


DY ( 


>UR 


3EN 


T L( 


)SS 






























-,n-C 


< 0K & 


S^ 






















60-X 








- 
















25 C 


YCLES 






'- 



.20 



.04 



3456789 10 
KILO-MAXWELLS PER SQ. CM. 
Fig. in. 



11 12 



curves are plotted from values calculated by means of the 
formula and refer to laminations usually employed, these 
being 0.014 i ncn thick. 

(b) Hysteresis loss. A certain amount of power, P h , 
due to the presence of hysteresis, is required to carry the 



158 



ALTERNATING-CURRENT MACHINES. 



iron through its cyclic changes. The value of P h can be 
calculated from the formula expressing Steinmetz's Law, 

P h = icr 7 vfv<B> m lm \ 
where 

v = volume of iron in cm. 3 , 
/ = frequency, 
<B m = the maximum flux density, 
and v = the hysteretic constant. 

A fair value of ?j for transformer sheets is .0021. Curves 
of the hysteresis loss in 25 cycle and 60 cycle transformers 
based upon this value of i) are shown in Fig. 112. In the 
better grades of transformers, however, the hysteresis loss 



3.0 



q2.5 

z 

O2.0 

Q. 
£l.5 

£l.O 
< 

$0.5 









i j j 
















H 


YST 


ERE 


SIS 


-OS! 










&/ 






















fe O 


^> 




























cS 






















J*£^ 



































,84 



.70 



56 



.42 



U 



3456789 10 
KILO-MAXWELLS PER SQ. CM. 

Fig. 112. 



11 12 



is less than that indicated by the curves by about 15 %. 
Hysteresis loss is practically independent of the load. 

In modern commercial transformers the core loss at 60^ 
may be about 75 % hysteresis and 25 % eddy current loss. 
At 125 ~ it may be about 60 % hysteresis and 40 % eddy 
current loss. This might be expected, since it was shown 
that the first power of / enters into the formula for hysteresis 
loss, while the second power of / enters into the formula 
for eddy current loss. 



THE TRANSFORMER. 1 59 

The core loss is also dependent upon the wave-form of 
the impressed E.M.F., a peaked wave giving a somewhat 
lower core loss than a flat wave. It is not uncommon to 
find alternators giving waves so peaked that transformers 
tested by current from them show from 5 % to 10 % less 
core loss than they would if tested by a true sine wave. 
On the other hand generators sometimes give waves so 
flat that the core loss will be greater than that obtained by 
the use of the sine wave. 

The magnitude of the core loss depends also upon the 
temperature of the iron. Both the hysteresis and eddy cur- 
rent losses decrease slightly as the temperature of the iron 
increases. In commercial transformers, a rise in tempera- 
ture of 40 C. will decrease the core loss from 5 % to 10 %. 
An accurate statement of the core loss thus requires that 
the conditions of temperature and wave-shape be specified. 

62. Exciting Current. — In commercial transformers the 
exciting current lags less than 90 behind the primary im- 
pressed E.M.F., because of the iron losses. The exciting 
current may therefore be resolved into two components, 
one in phase with the primary E.M.F., and the other at 
right angles to it. The former is that current necessary to 
overcome the core losses and is called the power component 
of the exciting current. It is expressed as 

T _ Pe+Pk 

the values of P e and P h being calculated from the formulas 
of § 61. 

The other component, being 90 behind E p , is termed the 
wattless component of the exciting current, or the magnetizing 



i6o 



ALTERNATING-CURRENT MACHINES. 



current of a transformer. It is that current which sets up the 
magnetic flux in the core, and is denoted by the symbol I mag . 
Representing the reluctance of the core by (ft, and the 
magnetomotive force necessary to produce the flux <£ m by 
X, from §§ 21 and 25, vol. L, 

L 

4 7in p 



<j> = — 



(ft 



nag 

IO 



whence 



10 (R<S> 



(ft 



io(R3>, 



4 ^^p 4V2 7tn p 
The value of (R is calculated (§ 24, vol. i.) from 



(ft 



/ 
I* 



where / is the length of magnetic circuit, A its cross-section 
and p the reluctivity of the iron 



p = - = 



:)■ 



permeability, 

The phase relations of the power and wattless components 
; k E of the exciting current are shown 

in Fig. 113. The angle between 
I exc and I mag is called the angle of 
hysteretic advance and is denoted 
by a. This angle is determined 
from the relation 




tan' 



-1 * e+h 

J- rnnn 



Fig - " 3 - It should be noted that the use 

of the term hysteretic in this connection is somewhat mis- 
leading, for the value of a depends upon the eddy cur- 



THE TRANSFORMER. l6l 

rent loss as well as upon the hysteresis loss. The exciting 
current is 



7_ . / 72 I 72 

and lags behind the primary impressed E.M.F. by an angle 
90 — a. 

The magnitude and position of the exciting current of a 
transformer can be determined experimentally by the use 
of a wattmeter, a voltmeter, and an ammeter connected in 
the primary circuit, the secondary, of course, being open- 
circuited. The ammeter reading gives the value of I exc and 
its position is given by the equation 

• -1 P 

a = sin — , 

J - jL exc 

P being the wattmeter reading minus the copper loss due to 
the exciting current in the primary winding. 

If the impressed E.M.F. be harmonic the flux will also 
be harmonic and consequently the magnetizing current 
cannot be harmonic, because of the variation in the reluc- 
tance of the core. Besides a decrease in permeability with 
increasing flux density, the permeability on rising flux is 
smaller than on falling, under a given magnetomotive force, 
due to hysteresis. Therefore the magnetizing current wave 
will be peaked and will have a hump on the rising side. 
This magnetizing current wave can be plotted when the 
hysteresis loop of the core is given over the range of flux 
density produced by the primary E.M.F. Since 5C is 
directly proportional to the current, and & is proportional 
to 3>, if proper units are chosen, 3C m may be taken equal 
to I mag m , and ($> m may be taken equal to & m . Then the 
hysteresis loop and the sinusoidal flux curve may be drawn 



1 62 



ALTERNATING-CURRENT MACHINES. 



as in Fig. 114. The value of the magnetizing current 
corresponding to a given value of the flux is obtained by 
taking the abscissa corresponding to this flux value from 
the hysteresis loop and laying it off as an ordinate at the 
point on the time axis corresponding to the flux value taken. 
This process is indicated in the figure, and the entire current 
curve has been constructed by proceeding in this manner. 

This distorted curve of magnetizing current may be 
resolved into true sine components (§ 10), a fundamental 
with higher harmonics, the third harmonic being the most 
pronounced. The exciting current, being composed of 




Fig. 114. 

two components, one of which is non-sinusoidal, will also be 
non-sinusoidal, but since it is usually very small compared 
with the load current, no appreciable error will be intro- 
duced by considering I exc as harmonic. 

The exciting current varies in magnitude with the design 
of the transformer. In general it will not exceed 5 % of the 
full load current, and in standard lighting transformers it 
may be as low as 1 %. In transformers designed with 



THE TRANSFORMER. 163 

joints in the magnetic circuit the magnitude of the exciting 
current is largely influenced by the character of the joints, 
being large if the joints are poorly constructed. 

63. Equivalent Resistance and Reactance of a Trans- 
former. — If a current of definite magnitude and lag be 
taken from the secondary of a transformer, a current of 
the same lag and v times that magnitude will flow in the 
primary, neglecting resistance, reluctance, and hysteresis. 
An impedance which, placed across the primary mains, 
would allow an exactly similar current to flow as this 
primary current, is -called an equivalent impedance, and its 
components are called equiva- 



#w\h 



Fig. 115. 



lent resistance and equivalent 
reactance. 

If the secondary winding of /■ 
the transformer have a resis- 
tance R s and a reactance X s , ■ £_ 

and if the load have a resist- 
ance R 2 and a reactance X 2 , 
then the current that will flow in the secondary circuit is 

/ Es 

V(R S +R 2 f + (X S + x 2 f 

where E s is the secondary induced pressure when E p is the 
primary impressed E.M.F. The secondary current lags 
behind E s by an angle <j> whose tangent is 

X s + X 2 



R s +R 



For convenience, X 2 and R 2 will be taken equal to zero, 
Fig. 115, and the expressions which will result will be the 



164 ALTERNATING^CURRENT MACHINES. 

equivalent resistance and reactance of the secondary wind- 
ing of the transformer. 

Therefore VR 2 + x) = ^- . 

If the equivalent impedance have a resistance R and a 

X X 

reactance X then the ratios and ■— must be equal, since 

R R s 

the angle of current lag is the same in both primary and 
secondary. And since the current in the equivalent im- 
pedance has the same magnitude as that in the primary 



and 

But 
and 



7 




E P 


lp 


Vr 2 


I,' 


/R 2 


+ x 2 - 


Ip 


= tIs, 




E p 


T 





E, 



therefore, VR 2 + X 2 = 4" = ^"F = ~* v ' R 2 + X s \ 



But 



t/ f T 2 1 

R = R, 
X X 



Solving R =— 2 R s , 



X — „ X s , 



which are the values of the equivalent resistance and 
reactance of the secondary winding respectively. Simi- 



THE TRANSFORMER. 165 

larly the equivalent load resistance and reactance are 
respectively 

R = ^ R 2 

and X = — X 2 . 

z 

64. Copper Losses. — The copper losses in a transformer 
are almost solely due to the regular current flowing 
through the coils. Eddy currents in the conductor are 
either negligible or considered together with the eddy cur- 
rents in the core. 

When the transformer has its secondary open-circuited 
the copper loss is merely that due to the exciting current 
in the primary coil, P exc R p . This is very small, much 
smaller than the core loss, for both I exc and R p are small 
quantities. When the transformer is regularly loaded the 
copper loss in watts may be expressed 

P c = ipRp + I S 2 R S , 
where R p and R s are the resistances of the primary and 
secondary coils respectively. In an ideal transformer 
with zero reluctance, I s is 180 behind I p , and this is also 
approximately true for a commercial transformer under a 
considerable load. Therefore, for convenience, the sec- 
ondary resistance may be reduced to the primary circuit 
and the copper loss may then be expressed as 

P c =i;(R p +R) =I P 2 (R P +^R S ). 

At full load this loss will considerably exceed the core 
loss. While the core loss is constant at all loads, the 
copper loss varies as the square of the load. 



66 



ALTERNATING-CURRENT MACHINES. 



65. Efficiency. — Since the efficiency of induction appa- 
ratus depends upon the wave-shape of E.M.F., it should be 
referred to a sine wave of E.M.F., except where expressly 
specified otherwise. The efficiency should be measured 
with non-inductive load, and at rated frequency, except 
where expressly specified otherwise. 

The efficiency of a transformer is expressed by the ratio 
of the net power output to the gross power input or by 
the ratio of the power output to the power output plus all 
the losses. The efficiency, e, may then be written, 

V.I, 

V S I S +P h +P e +P' 

where V s is the difference of potential at the secondary 
terminals. 

The losses and efficiencies of a line of 2200 volt, 60 cycle 
transformers of the shell type are given in the following 
table : — 



Rated. Output in 
Kilowatts. 


Core Loss in Watts . 


Full-load. Copper 
Loss in Watts. 


Per cent Efficiency 
at Full-load. 


I 


3° 


3 2 


94.1 


2 


5° 


56 


94.9 


3 


66 


78 


95-4 


5 


90 


i°5 


9 6 -3 


7-5 


116 


135 


96.8 


10 


135 


170 


97.0 


15 


169 


233 


97-4 


20 


200 


3°4 


97-5 


25 


225 


375 


97-65 


3° 


250 


444 


97.8 


40 


300 


586 


97-9 


5o 


35° 


7-'5 


98.0 



The efficiencies of a certain 10 K.w. transformer at 
various loads are shown by the curve of Fig. 116. 



THE TRANSFORMER. 



167 



If the transformer be artificially cooled, as many of the 
larger ones are, then to this denominator must be added 
the power required by the cooling device, as power con- 



100 


































90 

> 80 
O 
z 

uj 70 
O 

£6° 

H 
z 50 

Ui 

O 

06 in 






















































































lu 40 

CL 

30 
20 
10 



































































20 40 60 80 100 120 140 
PER CENT FULL LOAD 
Fig. 116. 

sumed by the blower in air-blast transformers, and power 
consumed by the motor-driven pumps in oil or water 
cooled transformers. Where the same cooling apparatus 
supplies a number of transformers or is installed to supply 
future additions, allowance should be made therefor. 

Inasmuch as the losses in a transformer are affected by 
the temperature, the efficiency can be accurately specified 
only by reference to some definite temperature, such as 

75° C. 

The all-day efficiency of a transformer is the ratio of 



168 ALTERNATING-CURRENT MACHINES. 

energy output to the energy input during the twenty-four 
hours. The usual conditions of practice will be met if the 
calculation is based on the assumption of five hours full- 
load and nineteen hours no-load in transformers used for 
ordinary lighting service. With a given limit to the first 
cost, the losses should be so adjusted as to give a maximum 
all-day efficiency. For instance, a transformer supplying 
a private residence with light will be loaded but a few 
hours each night. It should have relatively much copper 
and little iron. This will make the core losses, which con- 
tinue through the twenty-four hours, small, and the copper 
losses, which last but a few hours, comparatively large. 
Too much copper in a transformer, however, results in bad 
regulation. In the case of a transformer working all the 
time under load, there should be a greater proportion of 
iron, thus requiring less copper and giving less copper loss. 
This is desirable in that a loaded transformer has usually 
a much greater copper loss than core loss, and a halving 
of the former is profitably purchased even at the expense 
of doubling the latter. 

66. Calculation of Equivalent Leakage Inductance. — 

The magnetic leakage in a transformer is that flux which 
links with one winding and not with the other. Its mag- 
nitude depends upon the size and form of the coils and the 
manner of their arrangement. This magnetic leakage may 
be considered equivalent to an inductance connected in the 
primary circuit and to an inductance connected in the 
secondary circuit. After the leakage flux has been deter- 
mined, the inductances L p and L s are found from the 
relation 

I = n- , § 12 

i 



THE TRANSFORMER. 



169 



and then the reactances are obtained from X p = cuL p and 
X s = qjL s . 

To calculate the leakage flux, consider a shell type trans- 
former having one primary and one secondary coil with many 
turns of wire in each. The paths of the leakage flux in 
this type of transformer are indicated in Fig. 117. Let the 



COILS 




/,-; 



PRIM. 
=3= 





_ 




<©[! 




r 



K-P-*. KiSi">l 



Fig. 117. 

dimensions shown on the sections be expressed in centi- 
meters. 

It is convenient to consider the leakage flux as the sum 
of three portions, the part passing through the primary 
space, the part passing through the secondary space, and 
the part passing through the gap, g. The magnetomotive 
force tending to send flux through the elementary portion 

P 
dp and back through the iron is ~ of the whole M.M.F. of 

the primary, so for any element 



M.M.F. = 4 nn p i p 



I/O ALTERNATING-CURRENT MACHINES. 

where i p is expressed in absolute units. Since the per- 
meability of iron is roughly iooo times that of air, no 
appreciable error is introduced by considering the whole 
reluctance of the circuit of the leakage flux to be in the air 
portion of that circuit. The cross-section area of this air 
portion of the magnetic circuit for any element is 

- 2 dp = Xdp, 

2 

and its length is /, therefore the reluctance is — — . The 
elementary primary leakage flux, d® p , is then 

, M.M.F. _ 4 Tcn p i p p _ 4 nn p i p Xpdp 

p== 5T" "~ j_ ' p~~ ip ' 

Xdp 

Inductance, as defined in § 12, is numerically equal to 
the number of linkages per absolute unit of current, or 

l = n—- The number of turns linked with the elementary 

P 
flux d&p is -^ of the total number of primary turns, therefore 

the elementary leakage inductance dl p is 

d i -P n ^p -t n 4™ p i p Xpdp _ 4 7rn p 'X 
di *-p n * ip ~p U » l Pip ~ IP* ^ 

Integrating over the full width of the primary coil from 
o to P, there is obtained 

, _ 4 Tcn p 2 X f* p a2 ^ a _ 4 nKp^ P 



x -f p fd P = 



ip 2 j - c 1 3 

Proceeding in like manner, the value of l s is found to be 
j _ 4 nn s 2 X S 



THE TRANSFORMER. 171 

The leakage flux passing through the gap between the 
coils is set up by the entire magnetomotive forces of either 
the primary or the secondary. Consider the flux passing 
through the right-hand half of the gap to be set up by the 
secondary M.M.F., and that through the other half of the 
gap to be set up by the primary M.M.F. Then, proceeding 
as before, the leakage inductances equivalent to these 
portions of the leakage flux are respectively 

4 nn p 2 X g 
lffP ~ I ' 2 

and t-^.I. 

1 2 

Adding; the leakage inductances due to the primary and 
secondary M.M.F.'s are respectively, 



and l s + l as = 



lp+lffP ~ I {3 + 2) 



3 
4-n s H/S 



I \3 

Reducing to practical units, and multiplying by 10, the pri- 
mary and secondary leakage reactances are respectively 

4-^/P + l\ IO _ 9) (I) 

and Xj= l«pg + Q IO - 9) (2) 

all the dimensions being in centimeters. 

The secondary leakage reactance may be reduced to the 
primary circuit by dividing by r 2 (§ 63), but it should be 
remembered that this is only permissible when the trans- 
former is under considerable load or when the exciting 



172 



ALTERNATING-CURRENT MACHINES. 



current is entirely ignored, as in most practical calculations. 
The total equivalent leakage reactance in the primary 
circuit is then 



Atntal — 



4 7in p 2 coX / P 
~ V3 



H itrt 



(3) 



As some of the leakage flux passes through and between 
the coils where they project beyond the core, it is usual to 
take for X the mean length of a turn of a coil diminished by 
| of the length extending beyond the iron. 

The minimum leakage reactance would result if each 
secondary turn were immediately adjacent to a primary 
turn, but obviously this ideal condition cannot be attained 
in practice. Still it may be approximated by interleaving 
the secondary and primary coils. When one coil is placed 
between the two halves of the other, as in Fig. 141, the 
leakage reactance is approximately one fourth of that 
expressed by the foregoing formulae. The values assigned 
to the symbols for this case are indicated in Fig. 118. Thus, 



/ 




1 £ 


r 










\ 










f 












6 
<f> 


2 

DC 

a. 


6 

UJ 
CO 


■ 
\ 

1 
1 

i 










\ + 


i s 




4 8 










/ 



Fig. 118. 



by having many coils and by alternating primary and 
secondary coils, the leakage reactance may be greatly 
reduced. 



THE TRANSFORMER. 173 

The formulae for the calculation of leakage reactance 
may also be applied to the core type of transformers, but 
the notation will be slightly different. With this type, 
P and S are the radial depths of the primary and secondary 
coils respectively, g is the radial width of the gap, I is the 
axial length of the coils, X is the mean length of a turn of 
the windings, and n p and n s are the number of primary 
and secondary turns respectively on both sections. 

67. Regulation. — The definition of the regulation of a 
transformer as recommended by the American Institute of 
Electrical Engineers is as follows: "In constant-potential 
transformers, the regulation is the ratio of the rise of second- 
ary terminal voltage from rated non-inductive load to no- 
load (at constant primary impressed terminal voltage) to 
the secondary terminal voltage at rated load." Further 
conditions are that the frequency be kept constant, and 
that the wave of impressed E.M.F. be sinusoidal. 

Not the whole of the primary impressed E.M.F. is 
operative in producing secondary pressure, for I p R p volts 
are expended in overcoming the resistance of the primary 
coil, and I S R S volts are expended in overcoming the resist- 
ance of the secondary coil. In addition to these, a part of 
the impressed E.M.F. is lost in overcoming the primary 
and secondary reactances due to the leakage flux, the 
magnitudes of these decrements being I P X P and I S X S . 
To consider these various losses in voltage, imagine the 
transformer itself to be an ideal one, but to have a resist- 
ance R s , equal to the resistance of the secondary coil, and 
a reactance X s , equal to the secondary leakage reactance 
of the actual transformer, connected in the secondary 
circuit in series with the load resistance R 2 and reactance 



174 ALTERNATING-CURRENT MACHINES. 

X 2 . And further, let there be a resistance R p , equal to the 
resistance of the primary coil of the actual transformer, 
and a reactance X p , equal to the primary leakage reactance 
thereof, connected in the primary circuit of the ideal trans- 
former, as shown in Fig. 119. 

The complete vector diagram of E.M.F.'s and currents in 

\p ko c3 B i Y.. 



a transformer corresponding to the arrangement of Fig. 119 

is represented in Fig. 120, where 

V s = the difference of potential at the secondary terminals, 

E s = E.M.F. induced in secondary winding, 

E p = impressed primary pressure, 

E = operative part of E p , 

I p and I s = primary and secondary currents respectively. 

For clearness a 1 to 1 ratio has been portrayed, and the 

various drops are greatly exaggerated. The diagram will 

be discussed in detail. 

The exciting current, I exc , has two components, namely 
I mag in phase with the flux, and I e+h in phase with E . 
The magnetizing current is determined from the expression 

■ L mag ,— J s vz 

4 V 2 nAflflp 

and the power component of the exciting current is obtained 
from 

J-e+h — ^ • §01 

E p 



THE TRANSFORMER. 



175 



The current flowing in the secondary circuit is 



/. = 



V(X s +X 2 f + (R S + R 2 y 



§ 62 



and lags or leads the secondary induced E.M.F. by an 
whose tangent is 



angle 



X 



X. 



R, + R< 




the secondary induced elec- 
tromotive force being 90 
behind the flux. The pri- 
mary current, I p , is equal 
to the vectorial sum of — r 
times the secondary cur- 
rent and the exciting cur- 
rent as shown. When a 
small current is taken from 
the secondary of the trans- 
former, the directions of I p 
and I s are considerably less 
than 180 apart, but when 
the secondary current is 
large, the directions of I p 
and I s are approximately 
opposite. 

The secondary induced 
E.M.F. is not all utilizable 
at the terminals. There is a resistance drop of I S R S volts 
which is in phase with I s , and a reactance drop of I S X S 
volts due to the leakage flux, this being at right angles to 
the phase of the secondary current. The result of subtraot- 



Fig. 120. 



176 ALTERNATING-CURRENT MACHINES. 

ing I S R S and I S X S from E s vectorially is V s , which is the 
difference of potential at the secondary terminals. 

The operative part, E , of the primary impressed elec- 
tromotive force which is necessary to produce the secondary 

induced pressure E S} leads the latter by 180 and its mag- 

-p 
nitude is — L . There is a primary resistance drop of I P R P 

volts in phase with I p and a reactive drop due to leakage 
of I P X P volts at right angles to I p . Therefore the E.M.F. 
impressed upon the primary terminals necessary to produce 
E is the vectorial sum of E , I P R P and I p X p , and is denoted 
byE p . 

Both R s and R p become known quantities as soon as 
the size of the secondary and primary conductors is 
known. The values of X s and X p are calculated from 
the formulae derived in Art. 66. Thus all the quantities 
entering into the calculation of the vectors shown in Fig. 120 
are known. 

Then, when I s is the full-load current, the regulation of 
the transformer at power factor = cos cj> is 



E -I 

JL p 



Regulation = 



r _zEp~ V s 



V s V s 



Z 



which, when multiplied by 100, gives the percentage regu- 
lation. 

A circuit approximately equal to that of Fig. 119 is shown 
as Fig. 121, where the secondary resistances and reactances 
are reduced to the primary circuit, and where the exciting 
current is considered as flowing through a separate impe- 
dance, thus eliminating all transformer action. 



THE TRANSFORMER. 177 

A transformer diagram of practical importance is depend- 
ent upon the consideration that the exciting current may 
be neglected when the apparatus carries a large load. It 



72 



Ep 




JWWWWWV 






p 

Fig. i2i. 



follows that I p = tI s , and that I p is exactly opposite I s . 
The primary and secondary resistance drops, being in 
phase respectively with I p and I s , are parallel, and the 
latter may be reduced to the primary circuit and added 
algebraically to I P R P . Then the total equivalent resistance 

drop of the transformer is I p (r p + — jj. Similarly the 

total equivalent reactance drop of the transformer is 

I p ( X p H -* J and is at right angles to I p or I s . The 

impressed primary E.M.F., E p , is equal to the vectorial 

sum of E , I p (r p + -J-J, and I p (x p + -A as shown in 

Fig. 122. Hence the regulation is expressed by 

E x 



Ep tEL-K 



Regulation = = ^~ 



T 



In practice it will be found impossible to complete the 
solution of these diagrams graphically because of the 



i;8 



ALTERNATING-CURRENT MACHINES. 



extreme flatness of the triangles. The better way is to 
draw an exaggerated but clear diagram, and obtain the 

true values of the sides by 
the methods of trigonom- 
etry and geometry. 

The regulation of a trans- 
former at any load and 
power factor can be com- 
puted when the equivalent 
resistance and the equiva- 
lent reactance are known. 
The equivalent resistance 
can be determined experi- 
mentally rpy measuring the 
primary and secondary re- 
sistances using direct cur- 
rent, and then reducing the 
latter to the primary cir- 
cuit by dividing by t 2 . 
The equivalent reactance 
can be determined by 
short-circuiting one wind- 
ing and impressing a suffi- 
cient E.M.F. upon the 
other to permit full-load 
current to flow. This cur- 
rent value multiplied by 
the total equivalent resistance gives the resistance drop 
which must be subtracted from the impressed E.M.F. at 
the proper phase angle to obtain the total equivalent reac- 
tance drop. Dividing this by the current value there 
obtains the total equivalent reactance. The regulation of 




Fig. 122. 



THE TRANSFORMER. 



179 



the transformer at any load and power factor may there- 
after be calculated. 

This method is more reliable than the load test, in which 
the no-load and full-load voltages are directly measured, 
because of the magnitudes of these quantities. A slight 
error in these measurements would introduce a considerable 
error in the regulation value, for taking the difference 
between these large quantities exaggerates the error of 
measurement 



68. Circle Diagram. — The magnitude and phase of 

the current produced by a con- 
stant impressed primary elec- 
tromotive force, E p in Fig. 121, 
depends upon the resistance 
and reactance of the circuit. 
If the load be non-inductive, 
the current supplied to it is de- 
pendent upon the resistance of 
the load. Neglecting the effect 
of shunt exciting circuit, the 
impressed E.M.F. has two 
Fig - " 3 - components, that necessary to 

overcome the reactive drop due to the leakage flux in 
the transformer itself, and that necessary to overcome 
the resistance drop due to the resistance of the entire 
circuit. These are at right angles to each other and 

may be represented respectively by I p lX p + -y J and 

I p (r p + Rs + R *\ as in Fig. 123. If the resistance of the 
load be altered, the current will change and the point A 




i8o 



ALTERNATING-CURRENT MACHINES. 



X 

will be in a different position, since X p H — f is constant. 

However, the impressed E.M.F. is always equal and 
opposite to the resultant of the reactance drop and the 
resistance drop, and to satisfy this condition the locus of 
the point A must be a semicircle. As I p is proportional 
to the reactive drop, and since the two angles marked are 
equal, it follows that the locus of the point B is also a semi- 
circle. The diameter of this semicircle is 



X p + 3* 



T amperes, 



which is the condition corresponding to zero resistance. 
To sum up, then, the locus of the load current for various 
resistances, when the load is non-inductive, is a semicircle 
whose diameter is the ratio of the 
primary impressed E.M.F. to the 
total equivalent reactance of the 
transformer, and whose diameter 
is at right angles to E p . 

The total current produced by 
E p of Fig. 121, when the load is 



non-inductive (X 2 = o), is the 

vectorial sum of I p and I exc , as 

shown in Fig. 124. The resulting 

primary current lags behind E p by 

an angle (j> pf and the power factor 

of the complete circuit is the ratio 

of OM to ON, or cos <j> p . The 

power supplied to the transformer is the product of E p 

and OM, Knowing the copper and core losses, the out- 




THE TRANSFORMER. 



I8l 



put P may be computed, and the efficiency of the trans- 



former determined. 



The regulation is then obtained from 



Regulation 



P 

L 



rE p I s -P 




69. Methods of Connecting Transformers. — There are 
numerous methods of connecting transformers to distribut- 
ing circuits. The simplest case is that 
of a single transformer in a single- 
phase circuit. Fig. 125 shows such 
an arrangement. This and the suc- 
ceeding figures have the pressure and 
current values of the different parts 
marked on them, assuming in each 
Fi s- I2 5- case a i-k.w., i to 10 step-down trans- 

former. As in Fig. 126, two or more transformers may 
have their primaries in parallel 
on the same circuit, and have 
their secondaries independent. If 
the two secondaries of this case 
are connected properly in series 
a secondary system of double the 
potential will result, or by adding 
a third wire to the point of junc- 
ture, as shown by the dotted line 
of Fig. 127, a three-wire system 
of distribution can be secured. 
The secondaries must be connected cumulatively; that is, 
their instantaneous E.M.F.'s must be in the same direction. 
If connected differentially, there would be no pressure 



Fig. 126. 



182 



ALTERNATING-CURRENT MACHINES. 



T 






Fig. 127. 



between the two outside secondary wires, the instantane- 
ous pressures of the two coils being equal and opposed 
throughout the cycle. Again, 
with the same condition of pri- 
maries, the secondaries can be 
connected in multiple as in Fig. 
128. Here the connections must 
be such that at any instant the 
E.M.F.'s of the secondaries are 
toward the same distributing wire. 
The connection of more than two 
secondaries in series is not com- 
mon, but where a complex net- 
work of secondary distributing mains is fed at various 
points from a high-tension system, secondaries are neces- 
sarily put in multiple. 

In many types of modern transformers it is usual to 

wind the secondaries (low-ten- 
sion) in two separate and simi- 
lar coils, all four ends being 
brought outside of the case. 
This allows of connections to 
two-wire systems of either of 
two pressures, or for a three- 
wire system according to Figs. 
127 and 128, to be made with 
the one transformer, this being 
Fi s- I28 - more economical than using 

two transformers of half the size, both in first cost and 
in cost of operation. In many transformers the primary 
coils are also wound in two parts. In these, however, the 
four terminals are not always brought outside, but in some 



§ 




•* 




> 


20 A. 




f 




> 









;* 




* 






> 

> ° 







THE TRANSFORMER. 



183 



cases are led to a porcelain block on which are four screw- 
connectors and a pair of brass links, allowing the coils to 
be arranged in series or in multiple according to the pressure 
of the line to which they are to be connected. From this 
block two wires run through suitably bushed holes outside 
the case. 

A two-phase four-wire system can be considered as two 
independent single-phase systems, transformation being 

accomplished by putting similar 
single-phase transformers in the 
circuit, one on each phase. If 
it is desired to tap a two-phase 
circuit to supply a two-phase 
three-wire circuit, the arrange- 
ment of Fig. 129 is employed. 
By the reverse connections 
two-phase three-wire can be transformed to two-phase 
four-wire. An interesting transformer connection is that 
devised by Scott, which permits of transformation from 
two-phase four-wire to three-phase three-wire. Fig. 130 
shows the connections of the two transformers. If one 




Fig. 129. 





Fig. 130. Fig. 131. 

of the transformers has a ratio of 10 to 1 with a 
tap at the middle point of its secondary coil, the other 

must have a ratio of 10 to .867 I 10 to — ^J. One ter- 



1 84 



ALTERNATING-CURRENT MACHINES. 



minal of the secondary of the latter is connected to the 
middle of the former, the remaining three free terminals 
being connected respectively to the three-phase wires. In 
Fig. 131, considering the secondary coils only, let mn rep- 
resent the pressure generated in the first transformer. 
The pressure in the second transformer is at right angles 
(§7) to that in the first, and because of the manner of 
connection, proceeds from the center of mn. Therefore 
the line op represents in position, direction, and magnitude 
the pressure generated in the second. From the geo- 
metric conditions mnp is an equilateral triangle, and the 
pressures represented by the three sides are equal and at 
6o° with the others. This is suitable for supplying a 
three-phase system. In power transmission plants it is 
not uncommon to find the generators wound two-phase, 
and the step-up transformers arranged to feed a three- 
phase line. 

In America it is common to use one transformer for 
each phase of a three-phase circuit. The three transform- 
ers may be connected either 
Y or A. They may be Y on 
the primary and A on the 
secondary, or vice versa. 
Fig. 132 shows both primary 
and secondary connected A. 
The pressure on each pri- 
mary is 1000 volts, and as 
a i-k.w. transformer was Flg - I32, 

assumed, i.e., 1 k.w. per phase, there will be one ampere 
in each, calling for 1.7 (V3) amperes in each primary 
main (§ 45). This arrangement is most desirable where 
continuity of service is requisite, for one of the trans- 



-§< 



S_u 



1±1L 











k 17 A. 






sJ 


^<>' 








>?! 




> 






§ 




? 17 A. ■ 


Sog 







THE TRANSFORMER. 



I8 5 



formers may be cut out and the system still be opera- 
tive, the remaining transformers each taking up the dif- 
ference between J and h the full load; that is, if the 
system was running at full load, and one transformer was 

cut out, the other two 




Fig- 133- 



per cent. Even if two of 
them were cut out, ser- 
vice over the remaining 
phase could be main- 
tained. It is not uncom- 
mon to regularly supply 
motors from three-phase 
mains by two somewhat 
larger transformers rather than by three smaller ones. 
Fig. 133 shows the connections for both primaries and 
secondaries in Y. If in this arrangement one transformer 
be cut out, one wire of the system becomes idle, and only 
a reduced pressure can be maintained on the remaining 
phase. The advantage of the star connection lies in the 
fact that each transformer need be wound for only 57.7 
per cent of the line voltage. 
In high-tension transmis- 
sion this admits of build- 
ing the transformers much 
smaller than would be 
necessary if they were A 
connected. Fig. 134 shows 
the connections for prima- 
ries in A, secondaries in 
Y; and Fig. 135 those for primaries in Y and secondaries 
in A. By taking advantage of these last two arrange- 




Fig. 134. 



1 86 



ALTERNATING-CURRENT MACHINES. 



ments, it is possible to raise or lower the voltage with i 

to i transformers. With three i to i transformers, 

arranged as in Fig. 134, 100 

volts can be transformed to 

173 volts; while if connected as 

in Fig. 135, 100 volts can be 

transformed down to 58 volts. 

Fig. 136 shows a transformer 
and another one connected as 
an autotransformer doing the 
same work. Since the required 
ratio of transformation is 1 to 

2, the autotransformer does the work of the regular trans- 
former with one-half the first cost, one-half the losses, 
and one-half the drop in potential (regulation). The only 
objection to this method of transformation is that the pri- 
mary and secondary circuits are not separate. With the 
circuits grounded at certain points, there is danger that the 
insulation of the low-tension circuit may be subjected to 




Fig. 135- 




„-rA/wvwwwwvwwv- 

2000 v. I 



One 100 Kw. 
Transformer 
Ratio 1 to 2 



Losses not co_nsideired 



Uwwwj One 50 Kv*. 



I 10 



AAA/VJ 

1000 



Transformer 
o! Ratio 1 to 1 



Losses not considered. 
Fig. 136. 



the voltage of the high-tension circuit. One coil of an 
autotransformer must be wound for the lower voltage, and 
the other coil for the difference between the two voltages 
of transformation. The capacity of an autotransformer is 
found by multiplying the high-tension current by the dif- 
ference between the two operative voltages. Autotransfor- 
mers are often called compensators. Compensators are 



THE TRANSFORMER. 



I8 7 



advantageously used where it is desired to raise the poten- 
tial by a small amount, as in boosting pressure for very 
long feeders. Fig. 137 shows three 1 to 2 transformers 




Lo.ss.es .Q.ot considered 



Losses not considered 



Fig. 137- 



connected in A on a three-phase system, and three 1 to 1 
compensators connected in Y to do the same work. 

From a two-phase circuit, a single-phase E.M.F. of any 
desired magnitude and any desired phase-angle may be 
secured by means of suitable transformers, as shown in 
Fig. 138. Suppose the two phases X and Y of a two-phase 
system be of 100 volts pressure, and it is desired to obtain 
a single-phase E.M.F. of 1000 volts and leading the phase 
X by 30 . As in Fig. 139, draw a line representing the 





Fig. 138. 



DIRECTION OF PHASE X. 

Fig. 139- 



direction of phase X. At right angles thereto, draw a line 
representing the direction of phase Y. From their inter- 
section draw a line 1000 units long, making an angle of 



188 ALTERNATING-CURRENT MACHINES. 

30 with X. It represents in direction and in length the 
phase and the pressure of the required E.M.F. Resolve 
this line into components along X and Y, and it becomes 
evident that the secondary of the transformer connected 
to X must supply the secondary circuit with 866 volts, 
and that the secondary of the other must supply 500 volts. 
Therefore the transformer connected to X must step-up 
1 to 8.66 and that connected to Y must step-up 1 to 5. If 
10 amperes be the full load on the secondary circuit, the 
first transformer must have a capacity of 8.66 K.w., and the 
second a capacity of 5 k.w. The load on X and Y is not 
balanced. 

70. Lighting Transformers. — Because of the extensive 
use of transformers on distributing systems for electric 
lighting, the various manufacturers have to a great extent 
standardized their lines of lighting transformers. Some of 
these will be briefly described. 

The Wagner Electric Mfg. Co.'s "type M" transformer 
is illustrated in Fig. 140. It is of the shell type of con- 
struction, makers of this type claiming for it superiority 
of regulation and cool running. In the shell type the iron 
is cooler than the rest of the transformer, in the core type 
it is hotter. As the "ageing" of the iron, or the increase 
of hysteretic coefficient with time, is believed to be aggra- 
vated by heat, this is claimed as a point of superiority of 
the shell type. However, the prime object in keeping a 
transformer cool is not to save the iron, but to protect the 
insulation; and as the core type has less iron and generally 
less iron loss, the advantages do not seem to be remarkably 
in favor of either. In the Wagner "type M" transformers 
the usual practice of having two sets of primaries and sec- 



THE TRANSFORMER. 




Fig,. 140. 




Fig. 141. 



190 ALTERNATING-CURRENT MACHINES. 

ondaries is followed. Fig. 141 shows the three coils com- 
posing one set. A low-tension coil is situated between two 
high-tension coils, this arrangement being conducive to a 
good regulation. The ideal method would be to have the 
coils still more subdivided and interspersed, but practical 
reasons prohibit this. The space between the coils and 
the iron is left to facilitate the circulation of the oil in which 
they are submerged. The laminae for the shell are stamped 
each in two parts and assembled with joints staggered. 
As can be seen from the first cut, all the terminals of the 
two primary and the two secondary coils are brought outside 
the case. The smaller sizes of this line of transformers, 
those under 1.5 K.w., have sufficient area to allow their 
running without oil, so the manufacturers are enabled to 
fill the retaining case with an insulating compound which 
hardens on cooling. 

The General Electric Co.'s "H" transformers are of 
the core type. In Fig. 142 
is shown a sectional view 
giving a good idea of the 
arrangement of parts in 
this type. Fig. 103 is also 
one of this line of trans- 
formers. In it is shown 
the tablet board of porce- 
lain on which the connec- 
tions of the two high-ten- 
sion coils may be changed 
from series to parallel or Flg * I42 * 

vice versa, so that only two high-tension wires are brought 
through the case. Fig. 143 shows the arrangement of the 
various parts in the assembled apparatus. The makers 




THE TRANSFORMER. 



IQI 




claim for this type that the coils run cooler because of 
their being more thoroughly surrounded with oil than 
those of the shell type. Another point brought forward 
is that copper is a better conductor of heat than iron; the 

heat from the inner portions 
of the apparatus is more readily 
dissipated than in the shell 
type. The core has the advan- 
tage of being made up of simple 
rectangular punchings, and the 
disadvantage of having four 
instead of two joints in the 
magnetic circuit. A particular 
advantage of the "type H" 
transformer is the ease and 
certainty with which the pri- 
Flg ' I43 ' mary windings can be sepa- 

rated from the secondary windings. A properly formed 
seamless cylinder of fiber can be slipped over the inner 
winding and the outer one wound over it. This is much 
more secure than tape or other material that has to be 
wound on the coils. 

Fig. 144 shows a 2-K.w. O. D. transformer without the 
case. A tablet board is used for the terminals of the high- 
tension coils, but the low-tension wires are all run out of the 
case. Fig. 145 shows one of the coils. "Type O. D." trans- 
formers are built from \ to 25 k.w. for lighting and to 50 
K.w. for power. Those of 10 k.w. or less are in cast-iron 
cases, those above 10 k.w. in corrugated iron cases with 
cast tops and bottoms. The corrugations quite materially 
increase the radiating surface. The windings are sub- 
merged in oil. 



192 ALTERNATING-CURRENT MACHINES. 

An example of the Stanley Electric Manufacturing Co.'s 
standard line of "type A. 0." transformers is given in 




Fig. 144. 

Fig. 146. These are also of the shell type, with divided 
primaries and secondaries. 

71. Cooling of Transformers. — The use of oil to assist 
in the dissipation of the heat produced during the opera- 
tions of transformers is almost universal in sizes of less 
than about 100 k.w., especially if designed for outdoor 
use. Some small transformers are designed to be self- 
ventilating, taking air in at the bottom, which goes out at 
top as a result of being heated. They are not well pro- 
tected from the weather, and are liable to have the natural 
draft cut off by the building of insects' nests. Larger 



THE TRANSFORMER. 



193 




Fig. 145. 



transformers that are air cooled and that supply their own 
draft are used to some extent in central stations and other 




Fig. 146. 

places where they can be properly protected and attended 
to. A forced draft is, however, the more common. Where 
such transformers are employed, there are usually a number 



194 



ALTERNATING-CURRENT MACHINES. 



of them ; and they are all set up over a large chamber into 
which air is forced by a blower, as indicated in Fig. 147. 




000 

r&i s^ r^ 



1 „ 1 — r^-x — 1 „ 1 

AIR CHAMBER 



Fig. 147. 



Dampers regulate the flow of air through the transformers. 
They can be adjusted so that each transformer gets its 
proper share. 

Fig. 148 shows a General Electric Company's air-blast 
transformer in process of construction. The iron core is 
built up with spaces between the laminae at intervals .; and 
the coils, which are wound very thin, are assembled in 
small intermixed groups with air spaces maintained by 
pieces of insulation between them. The assembled struc- 
ture is subjected to heavy pressure, and is bound together 
to prevent the possibility of vibration in the coils due to 
the periodic tendency to repulsion between the primary and 
the secondary. These transformers are made in sizes from 
100 k.w. to 1000 k.w. and for pressures up to 35,000 volts. 

Another method of cooling a large oil transformer is to 
circulate the oil by means of a pump, passing it through a 
radiator where it can dissipate its heat. Again cold water 
is forced through coils of pipe in the transformer case, and 
it takes up the heat from the oil. There is the slight dan- 
ger in this method that the pipes may leak and the water 
may injure the insulation. Water-cooled transformers 
have been built-up to 2000 k.w. capacity. 



THE TRANSFORMER. 



195 



In those cases where the transformer requires some 
outside power for the operation of a blower or a pump, 
the power thus used must be charged against the trans- 




Fig. 148. 

former when calculating its efficiency. In general this 
power will be considerably less than 1 % of the trans- 
former capacity. 

72. Constant-Current Transformers. — For operating 
series arc-light circuits from constant potential alternating- 
current mains, a device called a constant-current trans- 
former is frequently employed. A sketch showing the 
principle of operation is given in Fig. 149. A primary 
coil is fixed relative to the core, while a secondary coil is 



1 96 



ALTERNATING-CURRENT MACHINES. 



allowed room to move from a close contact with the 
primary to a considerable distance from it. This secon- 
dary coil is nearly but not entirely 
counter-balanced. If no current 
is taken off the secondary that 
coil rests upon the primary. 
When, however, a current flows 
in the two coils there is a repul- 
sion between them. The counter- 
poise is so adjusted that there is 
an equilibrium when the current 
is at the proper value 




Fig. 149. 

If the current rises above this 
value the coil moves farther away, and there is an increased 
amount of leakage flux. 



This lowers the E.M.F. induced 




Fig. 150 



THE TRANSFORMER. 



197 



in the secondary, and the current falls to its normal value. 
Thus the transformer automatically delivers a constant 
current from its secondary when a constant potential is 
impressed on its primary. 

Fig. 150 shows the mechanism of such an apparatus as 
made by the General Electric Company. The cut is self- 
explanatory. Care is taken to have the leads to the mov- 




ing coil very flexible. Transformers for 50 lamps or 
more are made with two sets of coils, one primary coil 
being at the bottom, the other at the top. The moving 
coils are balanced one against the other, avoiding the 
necessity of a very heavy counterweight. Fig. 151 shows 
a 50-light constant-current transformer without its case. 
Fig. 152 shows a complete 2 5 -lamp apparatus. The tank 



198 ALTERNATING-CURRENT MACHINES. 

is filled with oil, the same as an ordinary transformer. 
Great care must be taken to keep these transformers level, 
and to assist in this the larger sizes have spirit-levels built 




Fig. 152. 

into the case. A pair of these transformers can be spe- 
cially wound and connected to supply a series arc-light 
circuit from a three-phase line, keeping a balanced load on 
the latter. 

73. Polyphase Transformers. — In transforming from 
one w-phase system to another w-phase system, instead of 
using n single-phase transformers, one ^-phase trans- 
former may be employed. A polyphase transformer con- 
sists of several single-phase transformers having portions of 
their magnetic circuits in common. As these common 
portions of the magnetic circuits carry fluxes differing in 



THE TRANSFORMER. 



199 



phase, an economy of material results due to the fact that 
the resultant flux is less than the arithmetical sum of the 
component fluxes. A further saving is effected due to the 



ihnrmr 



Fig- 153- 



necessity of only one instead of several containing cases, 
but this disappears, however, when the single-phase trans- 
formers are all mounted in one case. 

Three-phase transformers are used extensively in Europe 




Fig. 154. 

and the tendency toward their use in America is constantly 
increasing. The magnetic circuits of the two most common 
types of three-phase transformers are diagrammatically 



200 ALTERNATING-CURRENT MACHINES. 

shown in Fig. 153 and Fig. 154. The first is known as the 
core type, and the other as the shell type. 

The advantages claimed for the three-phase transformer 
over three single-phase units are: 1, a saving of about 15 % 
in first cost; 2, the required floor space is smaller and the 
weight is less; 3, greater efficiency. In the case of a break- 
down, however, the resulting derangement of the service 
and the cost of repair are greater for three-phase than for 
single-phase transformers. Another disadvantage is the 
greater cost of a spare unit. In large power stations an 
installation of three-phase transformers is believed to be 
more economical than an installation of single phase units. 

PROBLEMS. 

1. Determine the total flux of a 60 cycle lighting transformer having 
0.8 primary turn per volt of primary impressed E.M.F. 

2. Calculate the eddy current and hysteresis losses in the iron of a 
125 cycle core-type transformer for which $ m is 0.2 megamaxwell. The 
mean length of the magnetic circuit is 35 inches and the cross-section is 
9 square inches. 

3. If the E.M.F. impressed upon the primary of the transformer of 
the preceding problem be 2200 volts, compute the value of the exciting 
current and its phase. 

4. A transformer has 2000 turns of No. 16 B. & S. copper wire on the 
secondary winding, and 100 turns of No. 4 B. & S. copper wire on the 
primary winding. The mean lengths of the secondary and primary 
turns are respectively 17 and 28 inches. Determine the total equiva- 
lent primary resistance. 

.5. What is the copper loss in the transformer of the preceding problem 
when the primary current is 25 amperes, the exciting current being 
neglected ? 

6. Assuming that the transformer of problems 2 and 3 is a 5 K.w., 
20 : 1 step-down transformer, and that the primary and secondary 
resistances are 16.6 and .041 ohms respectively; determine the efficiency 
at full load. 



PROBLEMS. 201 

7. Find the all-day efficiency of the transformer of the preceding 
problem, basing the calculation upon 5 hours full-load and 19 hours 
no-load operation. 

8. Calculate the equivalent primary and the equivalent secondary 
leakage reactances of a 60 ~- shell type transformer having one primary 
and one secondary coil. The constants indicated in Fig. 117 are: 

Mean length of secondary turn 28 inches 
Mean length of primary turn 37.5 inches 
n p = 396 P = S = 1.25 inches 

n s = 18 g = .25 inch 

X = 15.5 inches / = 6.5 inches. 

What is the total equivalent primary leakage reactance, the transformer 
assumed to be under considerable load ? 

9. The resistances of the primary and secondary windings of a 1 : 2, 
60 cycle, step-up transformer are respectively 0.1 and .34 ohms. The 
equivalent leakage reactances of the primary and of the secondary are 
.14 and .5 ohms respectively, and the secondary induced electromotive 
force, E s , is 220 volts. Determine the E.M.F. to be impressed upon 
the primary terminals, the load on the secondary consisting of a 6 ohm 
resistance and an inductance of .01 henry. The exciting current is .85 
amperes and lags 70 behind E . 

10. Calculate the regulation of the transformer of the preceding 
problem. 

11. It is desired to transform from 2200 volts two-phase to 500 volts 
three-phase by means of a Scott transformer. Allowing one volt per 
turn on the windings, find the number of turns on each primary and on 
each secondary coil. 

12. A 100-1000 volt step-up transformer is connected to the A -phase 
of a two-phase, four-wire system, and a 100-2000 volt step-up trans- 
former is connected to the other phase. Determine the magnitude and 
phase of the secondary electromotive force when the secondary coils 
are in series. 



202 ALTERNATING-CURRENT MACHINES. 



CHAPTER VII. 

MOTORS. 
INDUCTION MOTORS. 

74. Rotating Field. — Suppose an iron frame, as in Fig. 
155, to be provided with inwardly projecting poles, and that 
these be divided into three groups, arranged as in the dia- 
gram, poles of the same group 
being marked by the same 
letter. If the poles of each 
group be alternately wound 
in opposite directions, and be 
connected to a single source 
of E.M.F., then the resulting current 
would magnetize the interior faces al- 
ternately north and south. If the im- 
pressed E.M.F. were alternating, then 
the polarity of each pole would change 
with each half cycle. If the three groups of windings 
be connected respectively with the three terminals of a 
three-phase supply circuit, any three successive poles will 
assume successively a maximum polarity of the same 
sign, the interval required to pass from one pole to its 
neighbor being one-third of the duration of a half cycle. 
The maximum intensity of either polarity is therefore 
passed from one pole to the next, and the result is a rotat- 
ing field. If the frequency of the supply E.M.F. be/, and 




MOTORS. 203 

if there be p pairs of poles per phase, then the field will 
make one complete revolution in ~ seconds. It will there- 
fore make - = — complete revolutions per second. A 
p 60 

rotating field can be obtained from any polyphase supply- 
circuit by making use of appropriate windings. 



75. The Induction Motor. — If a suitably mounted 
hollow conducting cylinder be placed inside a rotating field, 
it will have currents induced in it, due to the relative 
motion between it and the field whose flux cuts the surface 
of the cylinder. The currents in combination with the flux 
will react, and produce a rotation of the cylinder. As the 
current is not restrained as to the direction of its path, all 
of the force exerted between it and the field will not be in 
a tangential direction so as to be useful in producing rota- 
tion. This difficulty can be overcome by slotting the 
cylinder in a direction parallel with the axis of revolution. 
Nor will the torque exerted be as great as it would be if 
the cylinder were mounted upon a laminated iron core. 
Such a core would furnish a path of low reluctance for the 
flux between poles of opposite sign. The flux for a given 
magnetomotive force would thereby be greater, and the 
torque would be increased. 

Induction motors operate according to these principles. 
The stationary part of an induction motor is called the 
stator, and the moving part is called the rotor. It is com- 
mon practice to produce the rotating field by impressing 
E.M.F. upon the windings of the stator. There are, 
however, motors whose rotating fields are produced by the 
currents in the rotor windings. 



204 ALTERNATING-CURRENT MACHINES. 

The construction of a line of induction motors manu- 
factured by the General Electric Company is shown in 
Fig. 156. In this type the outer edges of the stator lami- 
nations are directly exposed to the air, thus improving 
ventilation. The stator core and windings of a Westing- 




Fig. 156. 

house induction motor are shown in Fig. 157. Each pro- 
jection of the core does not necessarily mean a pole; for 
it is customary to employ a distributed winding, there 
being several slots per pole per phase. The stator wind- 
ings are similar to the armature windings of polyphase 
alternators. The winding for each phase consists of a 



MOTORS. 



205 



group of coils, one group for each pole. The individual 
coils of each group are laid in separate slots. The stator 
windings of a three-phase induction motor are shown in 
Fig. 158, where each loop 
represents the group of coils 
for one pole. 

One type of rotor is shown 
in Fig. 159. The inductors 
are copper bars embedded in 
slots in the laminated steel 
core. They are all connected, 
in parallel, to copper collars or 
short-circuiting rings, one at 
each end of the rotor. They 
offer but a very small resist- 
ance, and the currents induced 

in them are forced to flow in a direction parallel with 
the axis. The reaction against the field flux is there- 




Fig- 157. 




fore in a proper direction to be most efficient in producing 
rotation. A rotor or armature of this type is called a 
squirrel cage. 



206 ALTERNATING-CURRENT MACHINES. 

Another type of rotor frequently used, especially in 
large induction motors, has polar windings which are 
similar to the windings on the stator. Fig. 160 shows a 




Fig. 159. 



rotor of this type made by the General Electric Company. 
The windings are short-circuited through an adjustable 
resistance carried on the rotor spider. When starting the 




Fig. 160. 

motor, all the resistance is in circuit, and after the proper 
speed has been attained, the resistance may be cut out by 
pushing a knob on the end of the shaft, as shown in the 
figure. This arrangement permits of a small starting 
current under load and a large torque, § 77. Fig. 161 



MOTORS. 



207 



shows another rotor of this type made by the same com- 
pany; the windings are identical with those on the other, 
except that their terminals are brought out to three slip- 




Fig. 161. 

rings. A starting resistance can be placed away from the 
motor and be connected with the rotor windings by means 
of brushes rubbing upon the slip-rings. 

76. Starting of Squirrel-Cage Motors. — To avoid the 
excessive rush of current which would result from connec- 
tion of a loaded squirrel-cage motor to a supply circuit, use 
is made by both the Westinghouse Company and the 
General Electric Company of starting compensators. 
These are auto-transformers which are connected between 
the supply mains, and which, through taps, furnish to the 
motor circuits currents at a lower voltage than that of 
the supply mains. After the rotor has attained the speed 
appropriate to the higher voltage, the motor connections 
are transferred to the mains, and the compensator is 
thrown out of circuit. The connections are shown in Fig. 
162, and the appearance of the General Electric Company 
compensator is shown in Fig. 163. The change of con- 



208 



ALTERNATING-CURRENT MACHINES. 



nections is accomplished by moving the handle shown at 
the right of the figure. While the compensator is supplied 
with various taps, only that one which is most suitable for 



Generator 




Running Side 



Oil Switch 



O-l O- 1 o-i o-J 



Starting Side 




Q 



fe4 



? 



^ 



Compensatoi~windinA 
Fig. 162. 

the work is used when once installed. The Westinghouse 
starter for squirrel-cage induction motors is shown in 
Fig. 164. It consists of auto-transformers and a multi- 
point drum-type switch, the latter being oil immersed so 
as to eliminate sparking at the points of contact. An 
important feature of this design is that the handle is moved 
in but one direction in passing from the off, through the 
starting, to the running position, thus making it impossible 
to connect the motor directly to the full line voltage. 

Where special step-down transformers are used for indi- 
vidual motors, or where several motors are located close to 
and operated from a bank of transformers, it is sometimes 
practical to bring out taps from the secondary winding, and 
use a double-throw motor switch, thereby making provision 



MOTORS. 



209 




Fig. 163. 




Fig. 164. 



210 ALTERNATING-CURRENT MACHINES. 

for starting the motor at low voltage, while avoiding the 
necessity for a starting compensator. 

The General Electric Company make small squirrel- 
cage motors, with centrifugal friction clutch pulleys; so 
that although a load may be belted to the motor, it is not 
applied to the rotor until the latter has reached a certain 
speed. The starting current is therefore a no-load starting 
current. 

77. Principle of Operation of the Induction Motor. — If 

the speed of rotation of the field be V r.p.m. and that of 

the rotor be V r.p.m., then the relative speed between a 

given inductor on the rotor and the rotating field will be 

V— V R.p.m. The ratio of this speed to that of the 

V— V 
field, viz., — — = s, is termed the slip, and is generally 

expressed as a per cent of the synchronous speed. If the 
flux from a single north pole of the stator be <3> maxwells, 
then the effective E.M.F. induced in a single rotor inductor 

is 2.22 p& s — - io~ 8 , where p represents the number of 
60 

pairs of stator poles. The frequency of this induced 
E.M.F. is different from that of the E.M.F. impressed 
upon the stator. It is 5 times the latter frequency. The 
frequency would be zero if the rotor revolved in synchro- 
nism with the field, and would be that of the field current 
if the rotor were stationary. As the slip of modern 
machines is but a few per cent (2 % to 15 %), the frequency 
of the E.M.F. in the rotor inductors, under operative con- 
ditions, is quite low. The current which will flow in a 
given inductor of a squirrel-cage rotor is difficult to deter- 
mine. All the inductors have E.M.F.' & in them, which at 



MOTORS. 211 

any instant are of different values, and in some of them the 
current may flow in opposition to the E.M.F. It can be 
seen, however, that the rotor impedance is very small. As 
the impedance is dependent upon the frequency, it will be 
larger when the rotor is at rest than when revolving. It 
will reduce to the simple resistance when the rotor is 
revolving in synchronism. Suppose a rotor to be running 
light without load. It will revolve but slightly slower than 
the revolving field, so that just enough E.M.F. is generated 
to produce such a current in the rotor inductors that the 
electrical power is equal to the losses due to friction, wind- 
age, and the core and copper losses of the rotor. If now a 
mechanical load be applied to the pulley of the rotor, the 
speed will drop, i.e., the slip will increase. The E.M.F. 
and current in the rotor will increase also, and the rotor 
will receive additional electrical power, equivalent to the 
increase in load. The induction motor operates in this 
respect like a shunt motor on a constant potential direct- 
current circuit. If the strength of the rotating field, which 
cuts the rotor inductors, were maintained constant, the slip, 
the rotor E.M.F., and the rotor current would vary directly 
as the mechanical torque exerted. If the rotor resistance 
were increased, the same torque would require an increase 
of slip to produce the increased E.M.F. necessary to send 
the same current, but the strict proportionality would be 
maintained. The rotating magnetism, which cuts the rotor 
inductors, does not, however, remain constant under vary- 
ing loads. As the slip increases, more and more of the 
stator flux passes between the stator and rotor windings, 
without linking them. This increase of magnetic leakage 
is due to the cross magnetizing action of the increased 
rotor currents. The decrease of linked field flux not only 



212 



ALTERNATING-CURRENT MACHINES. 



lessens the torque for the same rotor current, but alsc 
makes a greater slip necessary to produce the same cur- 
rent. The relation which exists between torque and slip 
for various rotor resistances is shown in Fig. 165, where 
the full lines represent torque, and the dotted lines current. 
An inspection of the curves shows that the maximum 



350,000- 


5-1-1 




££?""" x — ^ 


800,000- 






?^^ ^v / \ \ 


250,000- 


u^**""^"^ >^^^ \ \ 


O 


\;<> y% \ \ 


*- 200, 100- 
<o 




^SOs/ \ \\ 


z 


X. >C \\ \\ \ 


=» 


^•^^ N >» \ A \\ 1 


g 

{_ 150,000- 




£ 


XT ^\N \> \\\ 


100,000- 


^-- *- \n \ \\ 


60,000- 










STAND STILL 



SLIP# 

Fig. 165. 



SYNCHRONISM 



torque which a motor can give is the same for different 
rotor resistances. The speed of the rotor, however, when 
the motor is exerting this maximum torque, is different for 
different resistances. This fact is made use of in starting 
induction motors having wound rotors so that the starting 
current may not be excessive. The rotor resistance is 
designed to give full-load torque at starting with full load 
current. When the motor reaches its proper speed, this 
resistance is gradually cut out so that a large torque is 
secured within the operative range. 



MOTORS. 213 

78. Relation between Speed and Efficiency. — A portion 
of the total power supplied to the stator of an induction 
motor is consumed in the resistance of the stator windings, 
another portion is consumed in overcoming the stator iron 
losses, and the remainder is supplied to the rotor. 
Expressing this statement in the form of an equation, the 
total power supplied to the stator is 

P x = P h + P 2 , 

where P h is the sum of the stator copper and iron losses. 
Similarly, a portion of the power delivered to the rotor is 
consumed in heating the rotor windings, and another 
portion, very small and usually negligible, is consumed in 
overcoming the rotor iron losses, the rest being available 
as mechanical power. A small amount of the latter is 
wasted in bearing friction and windage, but this loss will 
be ignored. Then the power input to rotor is 

P = P 4- P 

where P C2 is the power expended in heating the rotor 
windings, and where P is the mechanical power developed 
in the rotor. Combining these expressions 

To obtain an approximate relation between efficiency and 
speed, it is convenient to neglect the stator losses; that is, 
the total power taken by the motor is considered as 
effective in producing the revolving field. Then the 
power input to the motor is 

P = P = P 4- P 

If the torque exerted by the rotating field upon the rotor 
be T lb. -ft., the power required to rotate this field at V 



214 ALTERNATING-CURRENT MACHINES. 

revolutions per minute against the reaction of the rotor 
will be 2 nVT ft. -lbs. per minute. The speed of the rotor 
being V revolutions per minute, the power developed in the 
rotor will therefore be 2izV'T ft. -lbs. per minute. The 
power expended in heating the rotor windings is 2 nVT 
— 2 nVT, or 2 tzT (V — V) ft. -lbs. per minute. Neglect- 
ing the stator losses and the rotor iron losses, the following 
proportion results: 

P t :P : P C2 = 2 nVT : 2 nV'T\ 2 tzT {V - V'\ 
or P x : P : P C2 = V : V : V - V = 1 : 1 - s : s ; 



from which 



P V 

Efficiency = — ^ = — - > 
Jr 1 V 



W = V (1 - *), P„ = P, (i - *), and P C2 = P x s. 

Thus the efficiency of an induction motor is approximately 
the ratio of the rotor speed to the synchronous speed, 
and the power expended in the rotor windings is approxi- 
mately proportional to the slip. Therefore, to secure a 
high efficiency, the slip should be small so that V more 
nearly approaches V; and to have a small slip requires a 
rotor having windings of low resistance. 

79. Determination of Torque. — The torque exerted by 
an induction motor may be expressed in terms of the 
stator input, stator losses, and synchronous speed. If P 
be the motor output in watts, then the torque in lbs. (one 
foot radius) is 

T = .33000 P n . 

2 1ZV 746 

But V = V (1 - s), and P = P 2 (1 - s). (§ 78) 
When P is expressed in watts, the term P 2 is the rotor 



MOTORS. 215 

input in synchronous watts, so-called. Therefore the 
torque, neglecting as usual the rotor iron losses, becomes 
= 33000 P 2 (1 - _£_) = K m 

2tzV (1 - s) 746 '' V 

The power which is delivered to the rotor is the differ- 
ence between the motor power intake and the stator losses, 

A. -f 

that is, P 2 = P 1 — P h ) and, since V = — — , the expres- 
sion for torque becomes 

P — P 
T =o.n 74 p ' r \ 

which is independent of rotor speed and mechanical out- 
put. 

80. The Transformer Method of Treatment. — It is cus- 
tomary in theoretical discussions to consider the induction 
motor as a transformer. Evidently when the rotor is 
stationary the machine is nothing but a transformer, with 
a magnetic, circuit so constructed as to have considerable 
magnetic leakage. When the rotor is moving, the machine 
does not act exactly like the ordinary transformer, but its 
action can be more conveniently and accurately determined 
by reference to transformer action. When no load is put 
upon the rotor of an induction motor, the currents supplied 
to the stator are called the exciting currents, just as is the 
current in the primary of a transformer when its secondary 
winding is open-circuited. The counter E.M.F.'s induced 
in the stator windings by the revolving flux is less than the 
impressed E.M.F.'s by amounts sufficient to allow the 
exciting currents to flow, and these overcome the eddy 
current and hysteresis losses of the stator iron, and set up 
the M.M.F.'s necessary to establish the rotating field. 



2l6 ALTERNATING-CURRENT MACHINES. 

When the induction motor operates under load, the 
slip, which before was practically zero, is increased, and 
E.M.F.'s are induced in the rotor windings due to the 
relative motion of the rotating field and the rotor. The 
demagnetizing effects of the rotor currents produced 
thereby are neutralized, as in the transformer with loaded 
secondary, by an increase of current in the stator windings, 
this being possible because of the diminished counter- 
electromotive forces. On account of the similarity of the 
actions of the induction motor and the transformer, the 
stator of machines as ordinarily constructed is also called 
the primary, and the rotor the secondary of an induction 
motor. 

When an induction motor is running at a certain slip, s, 
the frequency of the electromotive forces induced in the 
rotor windings is s-times the frequency of the supply 
voltage. Because of this fact, quantities in the secondary 
circuit cannot be directly added to quantities in the pri- 
mary circuit, but the reactions of the rotor currents and 
magnetic fluxes upon the primary are of the same fre- 
quency as the primary E.M.F.'s; for the flux produced by 
the secondary currents revolves relative to the rotor with a 
speed equal to the frequency of the induced secondary 
E.M.F.'s, so that the speed of this flux plus the speed of 
the rotor is the same as the speed of the revolving field. 
Thus, the secondary flux of an induction motor reacts 
upon the primary flux with the same frequency, exactly as 
in the transformer. 

81. Leakage Reactance of Induction Motors. — Not all 

of the flux set up by the stator currents traverses the air 
gap between the rotor and stator iron, nor does all of this 



MOTORS. 



217 



air gap flux link with the rotor turns, and, similarly, not 
all of the flux set up by the rotor currents links with the 
stator turns. The flux which links with one winding and 
not with the other is called the leakage flux. This 
magnetic leakage will be considered under the following 
heads: slot leakage, tooth-tip or "zig-zag" leakage, coil- 
end leakage, and belt leakage. 

Slot Leakage. The flux which passes across the slots 
and the slot openings is termed the slot leakage flux, the 







— «fe-» 






1 


/" 


- — 


/ /' _t 


^-- 


^ 


p-A ; n 


1 ' 1 

! 1 

I r~ 

I I 


[.":■ 


__v~_-._ 


■ w 

1 1 1 
1 1 
^ 1 1 1 

1 < 

11 


1 


ill! 






! ' 1 ! 






i'v 












— : 


y / 



Fig. 166. 



various paths thereof being shown by the dotted lines in 
Figs. 166 and 167. The magnitude of the slot leakage 
flux is dependent upon the form of the slot, and may be 
determined when the dimensions shown in the figures are 
known. Neglecting the reluctance of the iron portion of 
the magnetic circuit, the permeance of the path of the 
primary slot leakage flux per inch of slot length of the slot 
shown in Fig. 166 is 



0\ 



2-54 






+ - L + 



2C« 



+ 



W e 



w r 



w c 



218 



ALTERNATING-CURRENT MACHINES. 



where the subscripts i designate a primary slot. The 
method of derivation of this formula is identical with that 
given in § 66. 

The slot leakage flux per ampere inch of primary slot is 

$ 51 = 0.4 tt(P 1 » 1 , 

and hence the reactance of the primary slot leakage flux 
per phase in ohms is 

X S1 = 2 nfl s N^ <I> S1 io~ 8 , 

where l s is the length of slot in inches, n x is the number of 
conductors connected in series per 
primary slot, and N x is the num- 
ber of primary slots per phase. 

In motors having full-pitch wind- 
ings, all the conductors in one slot 
belong to the same phase wind- 
ing, but in motors having frac- 
tional-pitch windings some or all 
of the slots contain conductors 
belonging to different phase wind- 
ings, and therefore the conductors 
in these slots carry currents differing 
in phase. To take this influence 

into account, the pitch factor, C, must be inserted in the 

expression for slot leakage reactance. The value of C is 
coil span 
pole pitch 

expression for the equivalent reactance of the primary slot 




plotted in terms of 



in Fig. 168. Then the 



leakage flux per phase in ohms is 
X S1 - 2 CfhNrt \^~ + A. + 



2 c, 



w 8% + w c 



+ 



(I 



10 



w. 



(I) 



MOTORS. 



219 



Similarly, the reactance of the secondary slot leakage flux 
per phase in ohms is 

X S2 = 2 7iCfl s N 2 n 2 <& S2 io~\ 

In N \ 2 j) f 
which; when multiplied by ( 1 M ^, where p/ and p 2 are 

\fi 2 i\ 2 / p 2 

the number of primary and secondary phases respectively, 

reduces the secondary slot leakage reactance to the primary 

circuit, or 

N 2 p/[ a, 

1.0 



^\t~ 



™s 2 ™s 2 ™ S2 + W < 



W, 



]lO-'.(2) 



BE- 
O 

< 

"-.5 

I 

0. 

.3 



2 


^z2 


l/V 


M*~ 


c 2^ 


~7 


4 


/ 




c 



.1 .2 .3 .4 .5 .6 .7 .8 .9 ,1.0 
COIL PITCH 4- POLE PITCH 
Fig. 168. 

If the slots are of the form shown in Fig. 167, the per- 
meance of the elementary path dx per inch of slot length is 

d (r — r cos d) 
2 r sin 



dx 

2-54 — 

2y 



2.54 



1.27 dd. 



The leakage flux through this element per ampere inch 
of slot is 

Or 2 — r cos . r sin d~\ 



d® s = 0.4 



n (1.27 dd) U 



izr 



220 



ALTERNATING-CURRENT MACHINES. 



Hence the inductance per phase (considering only the cir- 
cular portions of the slots) is 

5 n 2 Nl s f n \d~ cos sin 6f dd, 

*J 



L< = 



71 IO 



and consequently the reactance of this portion of the slot 
leakage per phase in ohms is 

X s = 12.5 fn 2 Nl s io~ 8 = 2 fl s Nn 2 (0.625) ™~\ 

which is similar to the preceding equations for slot leakage. 
Hence equations (1) and (2) may be employed for calcu- 




Fig. 169. 

lating the slot leakage reactance of round slots, when the 
first three terms within the brackets of these expressions 
are replaced by the constant 0.625. 

Tooth-Tip Leakage. Tooth-tip leakage is that flux which 
passes through a portion of the tooth tip opposite a slot. 
The path of this leakage flux is shown in Fig. 169. For 
convenience in the following discussion, the number of 
rotor and stator slots will be assumed equal and their 
openings extremely small. The permeance of the path of 



MOTORS. 221 

this leakage flux is variable, being zero when a rotor slot is 
opposite a stator slot, and a maximum when a rotor slot is 
midway between two stator slots. The maximum per- 
meance per inch of slot length is 

I 

where X is the average or common tooth pitch, in inches, 
and A is the radial length of the air gap, in inches. The 
permeance of the path, when the rotor and stator are in an 

intermediate position, is 2.54 — • — — , and it follows 

that the average permeance per inch of slot length is 



2 -f±f~* (Xx-x>)dx = 0.423^ 



/ 



The tooth-tip leakage flux per ampere inch of slot for 
both stator and rotor is 

<$> t = 0.47^ = 0.532-^, 

and hence the equivalent reactance of the tooth-tip leakage 
per phase (stator and rotor) is 

X t = 2 nflsN^ $> t io~ 8 . 

When the number of slots in the rotor and in the stator 
is not the same, and when the slot openings are appreci- 

able, the value of $ t must be multiplied by ( 4- + -f — 1 ) , 

\X l X 2 ) 

where t x and t 2 are the equivalent stator and rotor tooth- 
tip widths respectively, and where X, and X 2 are the stator 
and rotor tooth pitches respectively. The equivalent 



222 



ALTERNATING-CURRENT MACHINES. 



tooth tips are determined by adding 2 A/ 7 to the actual 
tooth-tip width, where }' is the flux-fringing constant. 
The value of this constant is given in Fig. 170, where f is 



w r 



plotted in terms of — j* • Then introducing the pitch fac- 
tor, C, the expression for the equivalent reactance (stator 
and rotor) of the tooth-tip leakage flux per phase in ohms is 



x, = 3 . 3S c/yv 1 <-^ + - 






I IO 



(3) 



For motors having squirrel-cage rotors, the value of X t 
obtained from (3) should be reduced by 20 %. 

Coil-End Leakage. The flux which passes around the 







































/ 












/ 
















w 

"2A 







10 15 20 25 

Fig. 170. 




Fig. 171. 



ends of the coils where they project beyond the slots is 
called the coil-end leakage flux, the path of this flux being 
entirely or partly in air, as shown in Fig. 171. It is 
almost impossible to calculate accurately the coil-end 
leakage flux because of the proximity of the motor end 
plates and the influence of the mutual flux of the different 
phases. For a full-pitch three-phase winding, it is usual 



MOTORS. 223 

to assume this flux as one maxwell per ampere inch of 
exposed conductor. This assumption is experimentally 
justified. Then the flux in maxwells for all the con- 
ductors per pole per phase (i.e. per phase-belt) per ampere is 



<S> =Z ^ 






2 p 

where l c is the length of the end connections per primary 
turn, and p is the number of pairs of poles. 

The value of O c depends upon the ratio of the pole pitch 
to the diagonal of the section of the coil end, and is 
approximately proportional to the logarithm of this ratio. 
But the diagonal of the section of the coil end is approxi- 
mately equal to " - r — — = -7 , hence <I> C ispropor- 

number of phases p 

tional to log p'. For fractional pitch windings, the ratio 

■ — r— - = C r must be introduced. Therefore the value 

pole pitch 

of <E> C is proportional to log C p' . The value of log C p f 

for a full-pitch three-phase winding being 0.477, the 

inductance per phase-belt for any winding is therefore 

L e - l0 ^ ( n -^)\ xo-°. 

O.477 \ 2 p / 

If the length of the rotor coil-ends be considered 80 % of 
the stator coil-ends, then the total coil-end inductance per 
phase is 

L c = 3-78 i^~)p l c log C'p' . 10- 8 , 

and the coil-end leakage reactance per phase (stator and 
rotor) in ohms is 

X. = 5-95 / ^T^- 1, log C'f . xo- 8 . (4) 



224 



ALTERNATING-CURRENT MACHINES. 



Belt Leakage. Neglecting the exciting current of the 
induction motor, and considering an instant when a 
primary phase-belt completely overlaps a secondary phase- 
belt, the magnetomotive forces due to the currents in these 
belts of conductors will be in opposition, and there will be 
no belt leakage. But, if the belts of conductors be in 




Fig. 172. 



any other position, a secondary phase-belt overlaps two 
primary phase-belts, and their magnetomotive forces are 
no longer in opposition. The resultant M.M.F. will be 
effective in producing a leakage flux, termed belt leakage, 
which links with primary and secondary conductors, thus 
resulting in a leakage reactance. The relative position of 
stator and rotor for which this reactance is a maximum, is 
shown in Fig. 172, the paths of the flux being indicated by 
the dotted lines, and the slots for conductors of different 
phases being lettered differently. 

An expression for the average belt leakage inductance 
per phase, similar to that given by Adams, is 

£„ = k'hKK^CSJ ~io-", 



MOTORS. 



225 



where 5 12 is the number of series conductors per phase per 
pole (primary and secondary), D is the rotor diameter in 
inches, k' is 3.32 for two-phase and 1.005 f° r three-phase 

motors, K is the slot contraction factor, or — ^ , K 1 is a con- 

stant depending upon the number of slots per pole as 
obtained from Fig. 173, K 2 is a constant taking into 
account the ampere turns for the iron portions of the 





v\ 
















\\ 








































3E 




































12 16 20 24 

SLOTS PER POLE 



Fig- 173. 



belt leakage paths and may be taken as 0.85, and C is 
the pitch factor as determined from Fig. 168. The belt 
leakage reactance per phase (primary and secondary), in 
ohms, is 2 7ifL b , or 



X b = WKfiflgS^ — - . 



JA. 

X X X 2 



IO" 



(s) 



where k' f is 17.8 for two-phase and 5.36 for three-phase 
motors. This expression applies to induction motors 
having phase-wound rotors; for motors having squirrel- 
cage rotors the value of k" should be reduced by about 

65 %. 



226 



ALTERNATING-CURRENT MACHINES. 



Total Leakage Reactance. The total leakage reactance 
per phase of an induction motor is the sum of the various 
leakage reactances for which expressions have just been 
derived. That is, the total reactance per phase, X T , is 
equal to the sum of equations (i), (2), (3), (4), and (5). 

To secure a high starting torque and efficiency it is 





























^ 


*Z> 














































.6^-" 










\ 




UJ 

r> 
O 


















\ 




£T 
O 


















\\ 














sg*£ 


s£> 










\\ 


\ 






















\\ 






















\ 






















\ 



70 60 50 40 30 

PER CENT. SLIP 



Fig. 174. 

necessary to keep the magnetic leakage as small as possi- 
ble. The relation of torque to speed with various arbi- 
trary values of magnetic leakage is shown in Fig. 174. It 
is seen that the maximum torque increases directly as the 
leakage decreases. 

The leakage reactance of induction motors may be 
decreased by employing fractional-pitch windings, and by 
increasing the reluctance of the path of the leakage flux. 
The reluctance of the path of the useful flux, however, 



MOTORS. 227 

should be kept as low as possible, and it is usual to make 
the air gap just as small as is consistent with good mechani- 
cal clearance. Concentricity of rotor and stator is to be 
obtained by making the bearings in the form of end plates 
fastened to the stator frame. Some makers send wedge 
gap-gauges with their machines so that a customer may 
test for eccentricity due to wear of the bearings. A small 
air gap, besides lowering the leakage and raising the power 
factor, increases the efficiency and capacity of the motor. 

A convenient expression giving the proper radial depth 
of the air gap in inches, in terms of the horsepower rating 
of the motor, is 

80 

82. Calculation of Exciting Current. — The exciting 
current per phase of an induction motor has two com- 
ponents; one, the power component, which overcomes the 
eddy current and the hysteresis losses of the iron, the 
small stator copper loss due to the exciting current being 
neglected; and the other, the wattless component of the 
exciting current, which sets up the magnetomotive force 
necessary to overcome the reluctance of the magnetic 
circuit. The iron losses of the rotor under normal con- 
ditions are extremely small because of the very low fre- 
quency of rotor flux; therefore only the stator iron losses 
need be considered. In the following discussion, star- 
connected windings are assumed. 

The power component of the exciting current per phase is 

T P e + P h _ 

^=-^7-> §62. 



228 ALTERNATING-CURRENT MACHINES. 

where P e is the total eddy current loss, P h is the total 
hysteresis loss, E is the impressed electromotive force per 
stator winding, and p' is the number of phases. The 
values of P e and P h must be calculated separately for 
the stator teeth and for the stator yoke, because the flux 
densities in these parts of the magnetic circuit are different. 
The maximum flux density in the teeth will first be 
determined. 

The counter E.M.F. induced in a full-pitch distributed 
stator winding (one phase) by the rotating magnetic field is 

2 LLp^S ^-io- 8 , §42 

60 

where \ is the form factor or the ratio of the effective to 
the average E.M.F. , k 2 is the distribution constant as 
obtained from Fig. 57, p is the number of pairs of stator 
poles, <£ is the flux per pole and is assumed to be sinu- 
soidally distributed, 5 is the number of conductors con- 
nected in series per stator phase, and V is the rotor speed 
in revolutions per minute. In single-phase motors, the 
winding is usually distributed over two-thirds of the pole 
distance. When the rotor revolves at synchronous speed, 

then p — = /. If the ohmic drop due to the exciting 
60 

current in the stator winding be neglected, the counter 
E.M.F. and the impressed E.M.F. are practically equal. 
Finally, introducing the pitch factor, C, as obtained from 
Fig. 168, to take care of fractional-pitch windings, the value 
of the impressed electromotive force becomes 
E = 2 k&CQSf io" 8 , 
Eio s 



whence $ = 



2 kfoCSf 



MOTORS. 229 

Representing the core length in inches by l s , and the pole 
pitch in inches by l p , the average flux density (maxwells 
per square inch) in the air section becomes 

and the maximum flux density 



2 l s X p 

This equation assumes a continuous surface on both 
sides of the air gap over the polar region, but this does not 
occur in practice because of the presence of rotor and 
stator slots. In the following, the rotor slots are assumed 
extremely small so that their influence on flux distribution 
is inappreciable. If t t be the equivalent stator tooth tip 
width in inches, and X 1 be the stator tooth pitch in inches, 
then the maximum flux density (maxwells per square inch) 
in the air gap as well as in the tooth, which at that instant 
is at the center of the polar region, is 

<B«* = 7 1 <&» = A h * ; . ^ ' (1) 

which is the value to be taken for the maximum flux 
density in calculating the eddy current and hysteresis 
losses in the stator teeth. 

The maximum flux density in the stator yoke is half of 
the maximum flux density in the air gap where con- 
tinuous surfaces on both rotor and stator were assumed. 
Therefore the maximum flux density to be employed in 
calculating the eddy current and hysteresis losses in the 
stator yoke is 

/a n E iq8 ( \ 

my ~ Sk&Cl^Sf {2) 



230 ALTERNATING-CURRENT MACHINES. 

The values of P e and P h are calculated. as in § 61. 

The wattless component of the exciting current, or the 
magnetizing current, of an induction motor supplies the 
M.M.F. necessary to overcome the air-gap reluctance, 
the reluctance of the iron being neglected. The magneto- 
motive force required is 

M.M.F. = (ft <E = 2 A(B "' = ^ AE IO — , (3) 
2.54 S-oSW^W/ W 

where A is the radial length of the air gap in inches. 

In a two-phase machine, the magnetizing currents in 
both phases together set up this magnetomotive force. 
When the magnetizing current in one winding is at maxi- 
mum value {\^2 I mag ), at that instant the magnetizing 
current in the other winding is zero. Therefore the total 
M.M.F, set up per pole per phase will be 

M.M.F. = ■ 4^'SV2l nuV m } 

IO . 2 p W 

Equating (3) and (4) and solving, the magnetizing current 
per phase for a two-phase induction motor is 

j 5MAE10 8 

mag 5.08 V 2 \h 2 ci s i v t^f K5) 

In a three-phase machine, at the instant when the 
magnetizing current in one winding is at maximum value 
(V2 I m ag), the magnetizing current in each of the other 

two windings is at half maximum value ( 1 mag), and 

hence the current which sets up the M.M.F. is V2 I mag 

+ 2 ( I mag J = 2 V2 I mag , and the magnetomotive force 



MOTORS. 231 

supplied thereby, per pole per phase, is 

M.M.F. = 4 nS2 ^ 2 J ™* • ( 6 ) 

10 . 2 p v J 

Equating (3) and (6) and solving, the magnetizing current 
per phase for a three-phase induction motor is found to be 
one-half that for a two-phase motor, or half that given by 
equation (5). It should be noted that in the foregoing, 
E is the E.M.F. impressed upon each stator winding, and 
not the E.M.F. across motor terminals. 

After the two components of the exciting current have 
been calculated, the magnitude of I exc per phase may be 
obtained from the relation 

Iexc = Vl e l h + I mag J (7) 

and its angle of lag behind the impressed E.M.F. is given 
by 

& = cot-'^-- §62 

■*■ mag 

In induction motors the size of I e+h is small compared to 
Imag, the exciting current differing from the magnetizing 
current by but a few per cent. 

83. Circle Diagram by Calculation. — The similarity 
between an induction motor operating under a mechanical 
load and a transformer operating under a resistance load 
has already been pointed out; from whence it follows that 
the transformer circle diagram, § 68, may be applied to 
the induction motor. The circle diagram may be con- 
structed when the magnitude and phase of the exciting 
current are known, and when the leakage reactance of the 
motor has been calculated. The magnitude and position 



232 



ALTERNATING-CURRENT MACHINES. 



of the exciting current per phase may be computed from 
the expressions derived in the preceding article. This 
value is laid off as in Fig. 175 and the point D is thus 
located, which is one of the points on the circular current 
locus. The diameter of this circle is the ratio of the 
impressed E.M.F. per stator winding to the total leakage 

reactance per phase. That is, DC = — > where X T is 

the sum of equations (1), (2), (3), (4), and (5) of §81. 
Thus the circle diagram is completely determined. 

The current in a stator winding of an induction motor 




WATTLESS CURRENT 

Fig. 175- 



may be considered as the resultant of two components : one, 
the primary exciting current per phase, and the other, the 
effective current which supplies the magnetomotive force 
necessary to counterbalance the M.M.F. per phase due to 
the rotor currents. Thus in Fig. 175, OG is the stator 
current per phase, and is the resultant of OD, the exciting 
current per phase, and DG, the effective current per phase. 
The power factor of an induction motor depends upon 
the value of the stator current, as may be seen from the 



MOTORS. 233 

figure. The maximum power factor is attained when OG 
is tangent to the semicircle. Neglecting the power com- 
ponent of the exciting current, DB, the maximum power 
factor may be expressed as 

DC DC 

COS <j) m = 



AD+ D£ (TDC + D( L *<r+i 

2 2 

AD 

where <r = — — = leakage coefficient. The determination 

of the performance curves of an induction motor from the 
circle diagram will be considered later. 

84. Circle Diagram by Test. — The circle diagram of 
an induction motor is completely determined and may be 
constructed when the magnitude and position of the 
exciting current per phase and when the magnitude and 
corresponding position of any other current value per 
stator phase are known. 

The exciting current per stator phase can be determined 
when the primary amperes, the primary voltage between 
line wires, and the watts input have been obtained while 
the motor operates at no-load with full voltage. Thus, 
for a three-phase induction motor having its stator wind- 
ings Y-connected, Fig. 176, the magnitude of I exc per 
phase is the ammeter reading. The power input is the 
sum of the wattmeter readings in the two positions, that is, 
P x + P 2 = ^/~iE l I exc cos cf> e . § 47 

Therefore the angle by which I exc lags behind the im- 
pressed E.M.F. is 

1 P + P 
<f> e = cos -1 -fz L , 

v 3 EJ exc 



234 



ALTERNATING-CURRENT MACHINES. 



where E% is the voltmeter reading. The exciting current 
per phase, OD, Fig. 175, is now completely determined and 
may be drawn to a convenient scale. The point D con- 
stitutes one point on the circular current locus. 

If current, voltage, and power measurements be made 
when the motor is operating under load, or when the rotor 
is locked, another point, G, on the current locus may be 




Fig 176. 



similarly located. The latter measurement is to be pre- 
ferred because it determines the extreme value of the 
stator current, and intermediate points on the diagram are 
less likely to be in error. The impressed E.M.F., for the 
measurement with rotor locked, should be reduced to 
such a value as will send a current which will produce the 
same heating in the windings as does continuous full-load 
operation. The current and power may then be calculated 
for full voltage by increasing the former in direct pro- 
portion to the voltage, and by increasing the latter in 
proportion to the square of the voltage. The corrected 
current value is then laid off on the diagram to the same 
scale. The semicircle drawn through the points D and 



MOTORS. 235 

G, having a center on a line through D perpendicular to 

the E.M.F. vector, is the required current locus. 

Having constructed the circle diagram, the total leakage 

reactance per phase of the induction motor may be readily 

obtained, since the diameter of the semicircle is the ratio 

of the volts per stator phase to the total reactance per 

phase. The volts per stator phase in the three-phase 

Ei 
induction motor with Y-connected windings is — — • 

^3 

Before proceeding to the calculation of the performance 

curves of an induction motor, it is necessary to know the 
resistance per phase of the stator winding and the equiva- 
lent rotor resistance per stator phase. The resistance, R v 
between terminals of the stator winding may be directly 
measured by direct-current methods; thus for a three- 
phase motor, the total stator copper loss (§ 50) is f PR V 
where / is the current per line; hence the stator copper 

loss per phase is ■ — i, and the resistance per phase is 
2 

The equivalent rotor resistance per stator phase, r 2 , of a 
three-phase coil-wound rotor with a winding identical with 
that of the stator, is half the rotor resistance measured 
between two slip-rings. Therefore the copper loss of this 
coil-wound rotor per stator phase is Pr 2 . 

The equivalent resistance of a squirrel-cage rotor per 
stator phase cannot be determined directly, but may be 
calculated from the observations taken during the motor 
test with the rotor locked. As already explained, the 
impressed E.M.F. for this test is reduced, and the current 
and power input must thereafter be computed for full 



236 



ALTERNATING-CURRENT MACHINES. 



voltage operation. Representing the stator current per 
phase at lock by I L , the power per phase at lock by P L , 
and the constant iron and friction loss per phase, as 
obtained at no-load running, by P fi , then the copper loss 
in the squirrel-cage rotor of an induction motor per stator 
phase is 

P C2 = P L - P fi - I L \, 

and the equivalent resistance of the squirrel-cage rotor per 
stator phase is 

p 



85. Performance Curves from Circle Diagram. — Having 
constructed the circle diagram as described in the two 




Fig. 177. 



foregoing articles, the performance curves of the induction 
motor may be determined therefrom by calculating the fol- 
lowing quantities for a number of positions of the point G 
on the current locus, Fig. 177. 



MOTORS. 237 

Stator Current per phase = OG. 

Input to Motor = E.OK .p'. 

OK 
Power Factor of Motor = cos 6 = — 

Y OG 

Stator Copper Loss = r x . OG . p r . 

Input to Rotor = E . AK .p'-r t . (X? . p'. 

Rotor Copper Losses = r 2 . DG . p' . 

Motor Mechanical Output = 

p' [E . AK -r t .OG 2 - r 2 . DG 2 ]. 

Hi T^rc • Mechanical Output 

Motor Efficiency = - — 

y E.OK.p' 

Slip = 2 — = 2 • § 78 

E.AK - r,.OG 



rj, u E.AK. p'-r 1 .6G 2 .p' 
Torque = c.1174 p ^— x *— 



79 



In the foregoing, £ is the impressed E.M.F. per phase, 
p f is the number of phases, r x is the stator resistance per 
phase, r 2 is the equivalent rotor resistance per stator phase, 
p is the number of pairs of poles, and / is the frequency. 

The results obtained may then be embodied in a set of 
curves as in Fig. 178, where abscissae represent per cent 
full-load power output, and ordinates represent per cents. 

If the voltage impressed upon an induction motor be 
increased, there will result a proportional increase in the 
flux linked with the rotor, and in consequence a propor- 
tional increase in the rotor current. As the torque de- 
pends upon the product of the flux and the rotor ampere 
turns, it follows that the torque varies as the square of the 
impressed voltage. The capacity of a motor is therefore 



238 



ALTERNATING-CURRENT MACHINES. 



changed when it is operated on circuits of different volt- 
ages. 

Owing to the low power factor of induction motors, 
transformers intended to supply current for their operation 
should have a higher rated capacity than that of the 









































%s 


YNCH 


?ONO 


JS SP 


:ed 
































^ROJ 


QR^F 


EED_ 












A 


\y^ 






















^> 








f 


y 
























4 

o/ 
























-> 




/ 


7 








4/ 


























v c 






















40 
30 


/ 






^ 






















/ 






























/ 




























10 


1/ 





























40 60 80 100 

PERCENT FULL-LOAD POWER OUTPUT 

Fig. 178. 



120 



motors. It is customary to have the kilowatt capacity of 
the transformer equal to the horsepower capacity of the 
motor. 

The direction of rotation of a three-phase motor can be 
changed by transposing the supply connections to any two 
terminals of the motor. In the case of a two-phase, four- 
wire motor, the connections to either one of the phases 
may be transposed. 

86. Method of Test with Load. — The complete per- 
formance of two-phase or three-phase induction motors 



MOTORS. 239 

when operated from balanced two-phase or three-phase 
circuits may be calculated when the values of power input, 
as measured by the two-wattmeter method, § 47, have been 
determined by test for various mechanical loads upon the 
rotor. The instruments required are a voltmeter and a 
wattmeter, three observations being necessary at each 
load, namely, P v P 2 , and the line voltage, Ei. The pri- 
mary resistance measured between terminals and the 
equivalent secondary resistance per stator phase must be 
known. An outline of the method employed for a three- 
phase induction motor follows. 

By reference to § 47, it is seen that 

Total primary input = P ± + P 2 = V3 EJ cos cj>, (1) 

P 2 — P 1 = EJ sin <j), and (2) 

_ r ,_ P a - PJ , , 

Primary power factor = cos <j> = cos tan 1 v 3 „ p • (3 ) 

The primary current per terminal, as obtained from (2), is 

P — P 
/= £T^' (4) 



where sin (f> = sin tan 1 



^&t£1 : (s) 



The equivalent single-phase current is V3 /, §50. The 
power and wattless components of the equivalent single- 
phase current are respectively 

v^Jcos0 = ^-tA (6) 

and V3 / sin cf> = V3 P ' ~ P * ■ (7) 

The primary copper loss, § 50, is f PR V (8) 



240 ALTERNATING-CURRENT MACHINES. 

where P x is the stator resistance measured between ter- 
minals. The value of / is given by (4) and (5). 

The iron and friction loss is obtained by subtracting the 
primary copper loss at no-load from the total primary 
input at no-load, that is, 

Pt-.P' + P'-UfR,, ( 9 ) 

where P/, P 2 °, and I are the no-load values. I is 
obtained from (4) and (5) by taking P 2 ° for P 2 , and P x ° 
for P r 

The total primary losses including friction are therefore 

P fi + I PR V (10) 

The secondary input in synchronous watts is the differ- 
ence between the total primary input and the total primary 
losses, or 

P r = P x + P 2 -P fi -^PR,. (11) 

P P 
The external rotor torque, §79, is 0.1174 -y-> (12) 

where p is the number of pairs of poles and / is the fre- 
quency of the impressed E.M.F. 

The power and wattless components of the equivalent 
secondary current are respectively 

P 4- P —P° — P° 
I 2 cos j> 2 = (6) - (6 at no-load) = r ^ r * ^ t±> (l3 ) 

and 

/ 2 sin^ 2 =(7)-(7atno-load)=^[P 2 -P 1 -P 2 ° + P 1 ],(i4) 



hence the equivalent single-phase secondary current is 



/ = V(/ 2 cos^> 2 ) 2 + (/ 2 sin<£ 2 ) 2 . (15) 



MOTORS. 241 

The secondary copper loss = r 2 I 2 2 , where r 2 is the equiv- 
alent secondary resistance. 

The percentage rotor slip is the ratio of the secondary 
copper loss to the secondary input in synchronous watts, or 

100 rJ 2 



The output of the motor in horse power is 



(16) 



P - r I 2 

i#T' (I7) 

The efficiency of the induction motor, being the ratio of 
the watts output to the watts input, is 

P — r I 2 

E ff- = p +y ' ( i8 ) 

which, when multiplied by 100, gives the percentage 
efficiency. 

When numerous values of P x and P 2 have been experi- 
mentally determined and when the foregoing computations 
have been made for each set of readings, curves of the 
various factors may be plotted in terms of the motor out- 
put, as in the preceding article. 

87. Phase Splitters. — In order to operate polyphase 
induction motors upon single-phase circuits, use is made of 
inductances in series with one motor circuit to produce a 
lagging current, or of condensers to produce a leading cur- 
rent, or of both — one in each of two legs. The General 
Electric Company, in its condenser compensator, for use 
with small motors, as shown in Fig. 179, employs an 
autotransformer and condenser, connected as in diagram, 
Fig. 180. 



242 



ALTERNATING-CURRENT MACHINES. 



The autotransformer is used to step-up the voltage, 
which is impressed upon the condenser, to 500 volts. 




Fig. 179. 

The necessary size of the condenser is thereby reduced. 

The equivalent impedance of the autotransformer and 
condenser, as connected, is such as to pror 
duce a leading current in the one-phase 
sufficient to give a satisfactory starting 
torque, and it brings the power factor prac- 
tically up to unity at all loads. 

88. Single-Phase Induction Motors. — 

The difference between the single-phase 
and the polyphase induction motor lies 
principally in the character of their mag- 
netic fields. In the polyphase motor, the 
revolving field is practically sinuscidally 
distributed in space and constant in value. 
This is also true of the single-phase induc- 
tion motor when running at synchronous speed, but at any 
other speed the field is not constant in value, nor is it 




MOTORS. 



243 



sinusoidally distributed in space. If an alternating E.M.F. 
be impressed upon the stator winding of a single-phase 
induction motor an alternating flux will be set up which 
passes through the rotor. This pulsating flux lags approxi- 
mately 90 behind the impressed E.M.F. When the rotor 
is in motion, its conductors cut this flux and E.M.F. 's are set 
up in them which are in time phase with the flux. The value 
of these generated E.M.F. 's depends upon the magnitude 
of the stator flux and upon the speed of the rotor. They set 
up currents in the rotor conductors, the magnitudes of which 
are directly proportional to the electromotive forces generated 
therein. These currents set up a rotor flux which lags 90 
in time behind the rotor E.M.F.'s, and is displaced 90 elec- 
trical degrees in space. Thus the pulsating flux through the 
rotor due to the impressed alternating E.M.F. is at right 
angles, in a bipolar machine, to the rotor flux due to the 
motion of the rotor. When the stator flux is a maximum 
there will be no rotor flux, and when the stator flux is zero 
the rotor flux will be a maximum. In this way the resultant 
magnetic flux in the gap changes its position and revolves 
in the direction of rotation. At synchronous speed, the 
maximum values of the stator flux and the rotor flux are 
equal; thus a true rotating field of uniform intensity is pro- 
duced. At a lower speed, the maximum values are unequal 
and consequently the rotating field will be of variable inten- 
sity. At standstill no revolving field exists and no torque is 
developed. 

The inability of the single-phase induction motor to 
exert a torque at standstill has led to the introduction of 
numerous starting devices, but these are usually only 
applicable to small-sized motors. Three general methods 



244 ALTERNATING-CURRENT MACHINES. 

are employed to render the motor self -starting under load. 
First, the motor can be started as a repulsion motor 
(§103), and when normal speed is attained, a centrifugal 
device automatically short-circuits the commutator and 
simultaneously lifts off the brushes, thus changing the 
machine to a single-phase induction motor. Second, 
an auxiliary stator winding may be connected to the line 
through a phase-splitting device, as in § 87. Either a 
squirrel-cage or a coil-wound rotor may be used. Third, 
an auxiliary winding on the stator is connected through a 
non-inductive resistance and switching device to the line. 
An automatic clutch is employed, thus permitting the motor 
to approach normal speed before taking up its load. 

89. The Monocyclic System. — This is a system advo- 
cated by the General Electric Company for the use of 
plants whose load is chiefly lights, but which contains 
some motors. The monocyclic generator is a modified 
single-phase alternator. In addition to its regular winding, 
it has a so-called teaser winding, made of wire of suitable 
cross-section to carry the motor load, and with enough 
turns to produce a voltage one-fourth that of the regular 
winding, and lagging 90 in phase behind it. One end of 
the teaser winding is connected to the middle of the regu- 
lar winding, and the other end is connected through a slip- 
ring to a third line wire. 

A three-terminal induction motor is used, the terminals 
being connected to the line wires either directly or through 
transformers. 

90. Frequency Changers. — These are machines which 
are used to transform alternating currents of one frequency 



MOTORS. 245 

into those of another frequency. They are commonly used 
to transform from a low frequency (say from 25 or 40) to 
a higher one. They depend for their operation upon the 
variation with slip of the frequency of the rotor E.M.F.'s 
of an induction motor. The common practice for raising 
the frequency is to have a synchronous motor turn the 
rotor of an induction motor in a direction opposite to the 
direction of rotation of the latter's field. The synchronous 
motor and the stator windings of the induction motor are 
connected to the low-frequency supply mains. Slip-rings 
connected to the rotor windings of the induction motor 
supply current at the higher frequency. The size of the 
synchronous motor necessary to drive the frequency 
changer is the same percentage of the total output as the 
rise of frequency is to the higher frequency. 

91. Speed Regulation of Induction Motors. — The speed 
of an induction motor can be varied by altering the voltage 
impressed upon the stator, by altering the resistance of the 
rotor circuit, or by commutating the stator windings so as 
to alter the multipolarity. The first two methods depend 
for their operation upon the fact that, inasmuch as the 
motor torque is proportional to the product of the stator 
flux and the rotor current, for a given torque the product 
must be constant. Lessening the voltage impressed upon 
the stator lessens the flux, and also the rotor current, if the 
same speed be maintained. The speed, therefore, drops 
until enough E.M.F. is developed to send sufficient current 
to produce, in combination with the reduced flux, the 
equivalent torque. Increasing the resistance of the rotor 
circuit decreases the rotor current, and requires a drop in 
speed to restore its value. Both of these methods result 



246 



ALTERNATING-CURRENT MACHINES. 



in inefficient operation. If the impressed voltage be 
reduced, the capacity of the motor is reduced. In fact, the 
capacity varies as the square of the impressed voltage. 
Changes in the multipolarity of the stator require compli- 
cated commutating devices. 



92. The Induction Wattmeter. — The operation of the in- 
duction wattmeter, like that of the induction motor, is based 




Fig. 181. 

upon the action of a revolving or shifting magnetic field 
upon a metallic body capable of rotation. The rotating 
field is developed because of the difference in phase of the 
magnetic fields produced by the currents in the series and 
shunt coils of the wattmeter. The coils and the rotating 
member of an induction wattmeter are shown assembled in 
Fig. 181. The disk or armature is carried on a short 



MOTORS. 247 

shaft which is mounted in the usual way and is provided 
with a worm gear at its upper end for actuating the dial 
train. 

The series coil has no iron core and consists of a few 
turns of heavy wire, thus it possesses very little self-induction. 
If the power factor of the load circuit be unity, then the 
current flowing through this winding will be practically in 
phase with the impressed E.M.F., and, as the flux is in 
phase with the current producing it, this also will be 
approximately in phase with the impressed electromotive 
force. The shunt coil consists of a large number of turns 
of small wire wound on a laminated iron core. This 
winding has considerable self induction, hence the current 
flowing through it is almost at right angles to the impressed 
E.M.F. This angle of lag is slightly less than 90 owing 
to the iron and copper losses of the shunt circuit. The 
vector diagram corresponding to these conditions is shown 
in Fig. 182. 

The alternating magnetic fluxes due to the series and 
shunt coils pass through the disk and develop eddy cur- 
rents therein, which react on the fluxes and produce 
torque. As the torque is dependent upon both the flux of 
the series coil and that of the shunt coil, it is proportional 
to the energy which is to be measured. To render the 
angular velocity of the disk proportional to the torque, a 
permanent brake magnet is employed, and it is so mounted 
as to allow the disk to revolve between its poles. The 
permanent magnet may be moved radially with respect to 
the disk, and its position is adjusted to obtain the proper 
retarding force. 

It is necessary to have the series and shunt fluxes in 
time quadrature on non-inductive load in order that the 



248 



ALTERNATING-CURRENT MACHINES. 



wattmeter may indicate correctly on inductive load. To 
accomplish this, a copper band with a small gap in it, 
called a shading coil, is placed around the limb of the 
laminated iron core. This gap is closed by means of a 
resistance wire of such length and size that the E.M.F. 



B 

flser 



V-* 




*lsh 



Fig. 182. 



Fig. 183. 



induced in this band by the alternating shunt flux will 
send a current through it of such value that, when 
combined vectorially with the current in the shunt coil, 
a flux at right angles to the flux in the series coil will 
result. This is shown in Fig. 183, where (f> represents 
the angle by which the current in the series coil lags 
behind the impressed E.M.F. The vectors Esc and Isc 
represent respectively the electromotive force and the 
current in the shading coil. The resistance of the shading 
coil must be decreased when using the meter on circuits 
of lower frequency. 

Series transformers are used with induction wattmeters 
of more than 50 amperes capacity, and potential trans- 
formers are employed where the pressure exceeds 300 



MOTORS. 249 

volts. Polyphase induction wattmeters consist of separate 
single-phase elements assembled in the same case. The 
disks are mounted on a common shaft, and each revolves 
in its own field. 

SYNCHRONOUS MOTORS. 

93. Synchronous Motors. — Any excited single-phase 
or polyphase alternator, if brought up to speed, and if con- 
nected with a source of alternating E.M.F. of the same 
frequency and approximately the same pressure, will oper- 
ate as a motor. The speed of the rotor in revolutions 
per second will be the quotient of the frequency by the 
number of pairs of poles. This is called the synchronous 
speed; and the rotor, when it has this speed, is said to be 
running in synchronism. This exact speed will be main- 
tained throughout wide ranges of load upon the motor up 
to several times full-load capacity. 

To understand the action of the synchronous motor, 
suppose it to be supplied with current from a single 
generator. The following discussion refers to a single- 
phase motor, but may equally well be applied to the 
polyphase synchronous motor. In the latter case each 
phase is to be considered as a single-phase circuit. 

Let E x = E.M.F. of the generator, 

E 2 = E.M.F. of the motor at the time of connec- 
tion with the generator, 
6 = Phase angle between E x and E 2 , 
R = Resistance of generator armature, plus that of 
the connecting wires and of the motor 
armature, and 
ojL = Reactance of the above. 



250 ALTERNATING-CURRENT MACHINES. 

The resultant E.M.F., E, which is operative in sending 
current through the complete circuit, is found by combin- 
ing E t and E 2 with each 
other at a phase differ- 
ence 6, as in Fig. 184. 
Representing the angle 

•i^^H z x ^ between E x and E and 

K -^ E 2 and E by a and /? 

Flg " l84 " respectively, it follows that 

E = E x cos a + E 2 cos /?. 

This resulting E.M.F. sends through the circuit a current 
whose value is 

E 




VR 2 -\-u?L 2 
and it lags behind E by an angle <£, such that tan (f> = — • . 

The power P x which the generator gives to the circuit is 

P t = E t / cos (a - <j>) 

and the power P 2 which the motor gives to the circuit is 

P 2 = E 2 I cos (/? + 0). 

Now, if in either of the above expressions for power, the 
cosine has any other value than unity, then the power 
will consist of energy pulsations, there being four pulsa- 
tions per cycle. The energy is alternately given to and 
received from the circuit by the machine. If the cosine 
be positive, the amount of energy in one pulsation, which 
is given to the circuit, will exceed the amount in one 
of the received pulsations. The machine is then acting 
as a generator. If the cosine be negative the opposite 
takes place, and the machine operates as a motor. As a 



MOTORS. 251 

and p are but functions of E v E 2 , and 6, and as these latter 
are the quantities to be considered in operation, it is desir- 
able to eliminate the former. From the foregoing 

-P, = — l - cos (a — 6), 

1 VR 2 + oj'L 2 

or 

„ E? cos a + E X E 2 cos /? r , . . . ,, 

P v = — i * 2 [cos a cos 9 + sin a sin <p\. 

Zj 

Expanding, this becomes 

E 2 
P x = -^- (cos 2 a cos <f> + sin a cos a sin <f>) 

Zj 

-\ 1 — 2 - (cos a cos /? cos + sin a cos /? sin <£). 

Zj 

But 2 7T - = a + /?; 

hence cos a cos /? = cos (9 + sin a sin /?, 

and sin a cos /? = — sin (9 — cos a sin /?; 
also cos 2 a = 1 — sin 2 a. 

Therefore 

£ 2 
Pj = -±- (cos <£ — sin 2 a cos + sin a. cos a sin <j>) 

Zj 

EE 

+ ~^r 2 (cos Q cos </> + sin a sin /? cos <f> 

Zj 

— sin # sin cj> —cos a sin /? sin <£), 
or 

P 1 = [^f 2 - cos (S + <j>) + ^- COS tf] 

+ -— 1 [E 2 (sin a; sin /? cos <£ — cos a sin /? sin <f>) 
1 z 

— £ x (sin 2 a cos <£ — sin a cos a: sin <£)]. 



252 ALTERNATING-CURRENT MACHINES. 

But, since 

E, _ sin /? 
E 2 sin a 

the second term reduces to zero, and therefore 

P E A cos y + 0) + E i 2 _ cos 0. 

Vr 2 + osr Vr 2 + oj 2 l 2 

Similarly, the power supplied to the circuit by the motor is 

E,E ra i\ \ E» 

VR 2 +oj 2 L 2 Vr 2 



P 2= "7===== cos (^ - #) + -7==?==7 cos 



If there were no losses due to resistance, etc., P t would be 
numerically exactly equal to P 2 . Neglecting any losses 
in the machines, except that due to resistance, the alge- 
braic sum of P t and P 2 is equal to RP. In order to 
determine the behavior of a synchronous motor when on 
a given circuit, use is made of the above formula for power, 
and each case must be considered by itself. The method 
of procedure is shown in the next article. 

94. Special Case. — Suppose a single-phase synchronous 
motor, excited so as to generate 2100 volts, to be con- 
nected to a generator giving 2200 volts, the total resist- 
ance of the circuit being 2 ohms and the reactance 1 ohm. 
Then the angle <f> of current lag behind the resultant 

E.M.F. has a value tan <f> = — = 0.5, whence <j> = 26 34'. 

R 

Calculations of P 1 and P 2 for values of d between o° and 

360 have been made using the formulae of the preceding 

article, the results being embodied in the form of curves 

in Fig. 185. Phase differences, 6, are represented as 

abscissas and P x and P v in kilowatts, are represented as 



MOTORS, 



253 



ordinates. An enlargement of the lower portion of Fig. 185 
is shown in Fig. 186. The ratio of P 2 to P v when the 
former is negative and the latter positive, and when all 
losses excepting the copper losses are neglected, is the 
motor efficiency. From an inspection of these curves, and 































XT' 


X 


















/ 


\ 




\ 


\ 




















/ 


2800 


\ 


\ 


\ 


















/ 






\ 


\ 


\ 














/ 




1- 
V- 
< 
^ 2000 

O 

-1 




\ 




\ 










/ 




/ 








Vi 


\ 










/ 




1 




1200 
800 










\' 
























\ 


























\ 

















-400 










k^ 









































120 150 180 210 2i0 270 300 
6 DEGREES 

Fig. 185. 



a consideration of the equations from which the curves are 
derived, the following conclusions may be drawn : — 

(a) The motor will operate as such for values of 6 
between 175 and 23 8°. The difference between these 
angles may be termed the operative range. 

(b) The generator would operate as a motor for values 



254 



ALTERNATING-CURRENT MACHINES. 



of between 133 and 174 , providing the motor were 
mechanically driven so as to supply the current and power; 
i.e., what was previously the motor must now operate as 
a generator. 

(c) The motor, within its operative range, can absorb 
any amount of power between zero and a certain maxi- 
mum. To vary the amount of received power, the motor 




600 3 



Fig. 186. 



has but to slightly shift the phase of its E.M.F. in respect 
to the impressed E.M.F. , and then to resume running in 
synchronism. The sudden shift of phase under change 
of load is the fundamental means of power adjustment in 
the synchronous motor. It corresponds to change of slip 
in the induction motor, to change of speed in the shunt 
motor, and to change of magnetomotive force in the 
transformer. 

(d) For all values of the received power, except the 
maximum, there are two values of phase difference d. 
At one of these phase differences more current is required 



MOTORS. 255 

for the same power than at the other. The value of the 
current in either case can be calculated as follows : — 

Since P x -f P 2 = RP, 



4'- 



A 



R 

The values of I are plotted in the diagram. The efficiency 

P 

of transmission z = — *■ is also different for the two values 

of 0. It is also represented by a curve. 

If the phase alteration, produced by an added mechan- 
ical load on the motor, results in an increase of power 
received by the motor, the running is said to be stable. If, 
on the other hand, the increase of load produces a decrease 
of absorbed power, the running is unstable. 

(e) If for any reason the phase difference 6, between 
the E.M.F.'s of the motor and generator, be changed to a 
value without the operative range for the motor, the motor 
will cease to receive as much energy from the circuit as it 
gives back, and it will, therefore, fall out of step. Among 
the causes which may produce this result are sudden 
variations in the frequency of the generator, variations 
in the angular velocity of the generator, or excessive 
mechanical load applied to the motor. In slowing down, 
all possible values of will be successively assumed; and 
it may happen that the motor armature may receive suffi- 
cient energy at some value of to check its fall in speed, 
and restore it to synchronism, or it may come to a stand- 
still. 

(/) Under varying loads the inertia of the motor arma- 
ture plays an important part. The shifting from one 
value of d to another, which corresponds to a new median- 



256 ALTERNATING-CURRENT MACHINES. 

ical load, does not take place instantly. The new value is 
overreached, and there is an oscillation on both sides of 
its mean value. This oscillation about the synchronous 
speed is termed hunting. If the armature required no 
energy to accelerate or retard it, this would not take 
place. 

(g) The maximum negative value of P 2 — that is, the 

maximum load that the motor can carry — is evidently 

when cos (d — <f>) = — 1 or when — <j> = 180 . The 

formula for the power absorbed by the motor then reduces 

D E 2 cos <f> - E X E 2 

tO P 2m = 2 , 1 2 = 302 K.W. 

VR 2 +co 2 L 2 6 

(h) The operative range of the motor can be determined 
by making P 2 equal to zero. By transformation the for- 
mula then becomes 

COS (6 — <j>) = — — ^ COS (f). 

E l 

Two values of (0 — cf>) result, one on each side of 180 . 
In the case under consideration cos (0 — <j>) = —.8537, and 
6 - $ = 211 23' or 148 37 r . Since = 26 34', 6 = 237 
57' or 175 n r . 

95. The Motor E.M.F. — To determine what value of 
E 2 will give the maximum value of power to be absorbed 
by a motor, consider E 2 as a variable in the equation given 
in (g) above. 

Differentiating 

dP 2m _ 2 E 2 cos $ — E A 



dE 2 V R 2 + co 2 L 2 
and setting this equal to zero and solving, 

E = 1 — - =1230 volts. 

2 2 cos ^ ° 



MOTORS. 



257 




At this voltage the maximum possible intake of the motor 
is 605 k.w. If the voltage of the motor be above this or 
below it, its maximum intake will be smaller. 

Remembering that the current lags behind the resultant 
pressure of the generator and motor pressures by an angle 
(j>, which is solely dependent upon co, 
L> and R, it will be easily seen, from 
an inspection of Figs. 187, 188, and 
189, that the current may be made 
to lag behind, lead, or be in phase 
with E v by simply altering the value 
of E 2 . This may be done by vary- 
ing the motor's field excitation. A 
proper excitation can produce a unit 
power factor in the transmitting 
line. The over-excited synchronous 
motor, therefore, acts like a con- 
denser in producing a leading cur- 
rent, and can be made to neutralize the effect of induc- 
tance. The current which is consumed by the motor for 
a given load accordingly varies with the excitation. The 
relations between motor voltage and absorbed current for 
various loads are shown in Fig. 190. 

Synchronous motors are sometimes used for the purpose 
of regulating the phase relations of transmission lines. 
The excitation is varied to suit the conditions, and the 
motor is run without load. Under such circumstances the 
machines are termed synchronous compensators. 

The capacity of a synchronous motor is limited by its 
heating. If it is made to take a leading current in order 
to adjust the phase of a line current, it cannot carry its 
full motor load in addition without excessive heating. 



C ,,e "» c NT • ^DING E, 

Fig. 188. 



CURRENT IN PHASE WITH E, 

Fig. 189. 



258 



ALTERNATING-CURRENT MACHINES. 




Fig. 190. 



96. Starting Synchronous Motors. — Synchronous motors 
do not have sufficient torque at starting to satisfactorily come 
up to speed under load. They are, therefore, preferably 
brought up to synchronous speed by some auxiliary source 
of power. In the case of polyphase systems an induction 
motor is very satisfactory. Its capacity need be but tV that 
of the large motor. Fig. 191 shows a 750-K.w. quarter- 
phase General Electric motor with a small induction motor 
geared to the shaft for this purpose. This motor may be 
mechanically disconnected after synchronism is reached. 
Before connection of the synchronous motor to the mains 
it is necessary that the motor should not only be in syn- 
chronism, but should have its electromotive force at a 
difference of phase of about 180 with the impressed 
pressure. To determine both these points a simple device, 
known as a synchronizer, is employed. The simplest of 
these is the connection of incandescent lamps across a 
switch in the circuit of the generator and motor, as shown 
in Fig. 192. When the phase difference between the gen- 
erator and motor E.M.F.'s is zero, the lamps will be 



MOTORS. 



259 



brightest, and when the phase difference is 180 , the lamps 
will be dark. As the motor comes up to synchronous 
speed, the lamps become alternately bright and dark. As 
synchronism is approached, these alternations grow slower 
and finally become so slow as to permit closing of the 
main switch at an instant when the lamps are dark. 




Fig. 19] 



Instead of connecting the synchronizing device directly 
in the main circuit, it may be connected in series with the 
secondaries of two transformers, whose primaries are con- 
nected respectively across the generator and motor ter- 
minals, as shown in Fig. 193. With this arrangement, 
maximum brightness of the lamps may indicate that the 
generator and motor E.M.F.'s are either in phase or in 



260 



ALTERNATING-CURRENT MACHINES. 



opposition, according to the manner in which the trans- 
former connections are made. 

Another synchronizing device which is now extensively 
used is known as the synchroscope, and is shown in 




Fig. 192, 




Fig. 193. 



Fig. 194. The instrument is provided with a pointer which 
rotates at a speed proportional to the difference of the 
generator and motor frequencies, the direction of rotation 
showing which is the greater. Thus, if the motor fre- 
quency is too high, the pointer will rotate anti-clockwise. 
When the frequencies are identical, the pointer assumes 
some position on the scale, and when this position coin- 
cides with the index at the top of the scale, the main 



MOTORS. 26l 

switch may be closed, thus connecting the two machines 
together. 

Synchronous motors may be brought up to speed with- 
out any auxiliary source of power. The field circuits are 
left open, and the armature is connected either to the full 
pressure of the supply, or to this pressure reduced by 
means of a starting compensator, such as was described 
in §76. The magnetizing effect of the armature ampere 




Fig. 194. 

turns sets up a flux in the poles sufficient to supply a small 
starting torque. When synchronism is nearly attained, the 
fields may be excited and the motor will come into step. 
The load is afterwards applied to the motor through 
friction clutches or other devices. There is great danger of 
perforating the insulation of the field coils when starting 
in this manner. This is because of the high voltage 
produced in them by the varying flux. In such cases 



262 ALTERNATING-CURRENT MACHINES. 

each field spool is customarily open-circuited on starting. 
Switches which are designed to accomplish this purpose are 
called break-up switches. 

97. Parallel Running of Alternators. — Any two alter- 
nators adjusted to have the same E.M.F. and the same 
frequency may be synchronized and run in parallel. 
Machines of low armature reactance have large synchro- 
nizing power, but may give rise to heavy cross currents, if 
thrown out of step by accident. The contrary is true of 
machines having large armature reactance. Cross cur- 
rents due to differences of wave-form are also reduced by 
large armature reactance. The electrical load is distributed 
between the two machines according to the power which is 
being furnished by the prime movers. This is accom- 
plished, as in the case of the synchronous motor, by a 
slight shift of phase between the E.M.F.'s of the two 
machines. The difficulties which have been experienced in 
the parallel running of alternators have almost invariably 
been due to bad regulation of the speed of the prime 
mover. Trouble may arise from the electrical side if the 
alternators are designed with a large number of poles. 
Composite wound alternators should have their series com- 
pounding coils connected to equalizing bus bars, the same 
as compound wound direct-current generators. 

SINGLE-PHASE COMMUTATOR MOTORS. 

98. Single-Phase Commutator Motors. — If the current 
in both field winding and armature of any direct-current 
motor be periodically reversed, the direction of rotation of 
the armature will remain unchanged. Therefore direct- 
current motors might be operated on alternating-current 



MOTORS. 263 

circuits. Shunt motors cannot be operated satisfactorily 
when fed with alternating current, because the reversals 
of current do not take place simultaneously in armature 
and field windings owing to the high inductance of the 
latter winding. This would cause momentary currents in 
the armature in a reversed direction and would tend to 
produce a counter-torque, thus considerably decreasing the 
effective torque. 

When direct- current series motors are supplied with 
alternating current, the instantaneous current value is 




Fig- 195- 

necessarily the same in both armature and field winding, 
and therefore no counter-torque is developed. The direct- 
current series motor with various modifications may be 
operated on alternating-current circuits, and when so used 
is termed the single-phase series motor, or the single-phase 
commutator motor. It is essential that the entire magnetic 
circuits of motors of this type be laminated in order to 
decrease the otherwise excessive hysteresis and eddy cur- 
rent losses. Series motors, when operated on alternating 
current, produce a pulsating torque varying from zero to a 
certain maximum value. 

The armature of a single- phase series motor is similar 
to that of the direct-current motor. The armature of a 
150 horse-power single- phase alternating-current railway 
motor is represented in Fig. 195. 



264 



ALTERNATING-CURRENT MACHINES. 



99. Plain Series Motor. — Consider a direct-current 
armature mounted within a single-phase alternating mag- 
netic field, as in Fig. 196. When the armature is station- 
ary an electromotive force will be induced in the armature 
turns, due to the alternating flux which passes between the 




Fig. 196. 



field poles. The greatest E.M.F.'s will be induced in the 
turns perpendicular to the field axis, since these turns 
link with the greatest number of lines of force; and no 
E.M.F.'s will be induced in the turns in line with the 
field axis. The directions of the E.M.F.'s induced in the 
armature turns by the change in field flux are indicated in 
the figure by the full arrows, and it is seen that the maxi- 
mum value of this E.M.F. is across BC. As in trans- 
formers, the effective value of this electromotive force 
(§ 59) is 

V2 IO 8 



Er = 



(1) 



MOTORS. 265 

where $ m is the maximum value of the flux entering the 
armature and N is the equivalent number of armature 
turns. 

The maximum number of lines of force linked with a 
single turn depends upon the position of this turn in the 
magnetic field, and is proportional to the greatest value of 
$ m times the cosine of the angle of displacement of the 
turn from the position AD. Assuming the turns to be 
evenly distributed over the periphery of the armature, the 
average value of the maximum flux linked with the arma- 

ture turns will be - <i> m . If there are N a conductors on 

TC 

the armature, the number of turns connected in con- 

N 
tinuous series will be — a • The electromotive forces 

2 

induced in the upper and lower groups of armature turns 
are added in parallel, consequently the effective number of 

• I N a N a rr^i r -, 

turns in series is — • = Therefore the equiva- 

224 ^ 

lent number of armature turns may be expressed as 

71 4 2 TZ 

Substituting this value of N in equation (1), the E.M.F. 
induced in the armature winding by the change in value of 
the field flux is 

77 J^m-t-V a 



V 



2 IO 



(3) 



and it lags 90 behind field flux in time. 

If the brushes of the motor, A and D, are placed at the 
points shown in Fig. 196, this electromotive force will not 



266 ALTERNATING-CURRENT MACHINES. 

manifest itself externally, since it consists of two equal 
and opposite components directed toward these brushes. 
This E.M.F. appears, however, in the coils short-circuited 
by the brushes, as will be shown later. The current, 
which enters the armature by way of the brush and which 
traverses the two halves of its windings in parallel, pro- 
duces an armature flux of maximum value & am . This sets 
up a reactance E.M.F. in the armature which in the case 
of uniform gap reluctance can be similarly expressed as 

K = — - - • U; 

V 2 IO 8 

This lags 90 behind the current. 

When the armature revolves, there are, in addition, 
electromotive forces induced in the armature conductors as 
a result of their cutting the field flux. The directions of 
these E.M.F. ,'s are indicated by the dotted arrows, and it 
is seen that these E.M.F. ,'s, generated by the rotation of 
the armature, add to each other and appear on the com- 
mutator as a maximum across AD. 

The average value of the electromotive force due to the 
rotation of the armature is 

F ro tav = <$>fN a — IO" 8 , § 42 

where V is the armature speed in rev. per min. in a 
bipolar field, and §> f is the field flux; and the effective 
value of this E.M.F. is 

(b f N V 

Er °'-V2lO* 60' (5) 

and is in time phase with the field flux, but appears as a 
counter E.M.F. at the brushes AD. 



MOTORS. 



267 



When an alternating current is passed through the field 
coils, the alternating field flux is set up, and this flux pro- 
duces a reactive E.M.F. in the field winding lagging 90 
behind the flux in phase, exactly as in a choke coil. The 
magnitude of this E.M.F. is 

V2 io 8 



(6) 



where <J> /m is the maximum value of the field flux, and N f 
is the number of field turns. 

The electromotive force, E, which is impressed upon the 
motor terminals, is equal and opposite to the vectorial 




Fig. 197. 

sum of E a , E rot , E f , and the IR drop of the armature 
and field windings, as shown in Fig. 197, where / is the 
current flowing through the field and armature, and <£ 
represents the phase of the flux. In this diagram, eddy 
current and hysteresis losses are ignored. The impressed 
electromotive force is therefore 

E = V(E rol + IRf + (E. + £,)'. (7) 



268 ALTERNATING-CURRENT MACHINES. 

ioo. Characteristics of the Plain Series Motor. — In the 

series motor, the same current passes through field and 
armature windings, and, if uniform reluctance around the 
air gap be assumed, then the armature and field fluxes will 
be proportional to the equivalent armature turns and field 
turns respectively. Therefore 

^ aa :^ /m = N:N / = — a :N}. (i) 

Representing the ratio of the field turns to the effective 

N 
armature turns by t, then <£ /m = r^ am , and N,= t ■ — a . 

2 1Z 

Therefore expressions (4) and (6) of § 99 become 
respectively 

F QjtoNg f 

£L a = = ■ • > 

V2 IO 8 * 

and E, = %^ • ft. 

V 2 IO 8 

Equation (5 ) of § 99 is 

V 2 IO 8 OO 

which then reduces to 

t V 1 V 

Erot — "7 E a —> and E rot = — E f -- • 

/ 60 A 60 

Therefore E f = r 2 ^. 

Neglecting the armature and field resistance drop, the 
impressed E.M.F. becomes 



/(£$ + (^ + i) 2 , (2) 



POWER TRANSMISSION. 269 

which is the fundamental E.M.F. equation of the plain 
series motor. 

The power factor of the motor is 

cos0= — = , (3) 



W®'+ <?+* 



and the current supplied to the motor is 

7 E E 

still neglecting the motor resistance. 

When V= 6of, the motor is said to run at synchronous 
speed (bipolar field). The power factor of a plain series 

motor, having r = 1, when running at this speed, is — =, 

or 0.446, and for values of r other than unity the power 

factor is less than 0.446. It is true that if the resistance 

of the motor be considered, the power factor will exceed 

this value, but nevertheless it remains extremely low. 

■p 
The current intake under these same conditions is — — — - . 

When the motor is at standstill, V= o, and the power 
factor is zero. The current intake at standstill is 

2X a 

Hence the ratio of the current at synchronism to the cur- 
rent at standstill is — = -f- - = 0.894. The ratio of the 

torque at synchronous speed to the torque at standstill, 
since it varies as the square of the current, is ( — =) -f- 



270 ALTERNATING-CURRENT MACHINES. 

(-) =0.80, which shows that the starting torque is but 

little greater than the torque at synchronous speed. For 
railway service motors are required having large starting 
torque and whose torque rapidly decreases as the speed 
of the motor increases. It is seen, therefore, that inde- 
pendent of its low power factor, the plain series motor, 
having uniform magnetic reluctance around the air gap, 
is unsuitable for traction and for similar purposes. 

If, however, the reluctance of the air gap in the direction 
AD, Fig. 196, is increased, the power factor and speed- 
torque characteristics will be improved, and these will 
depend largely upon the ratio of field turns to effective 
armature turns, as will be seen by considering the con- 
struction of the motor to be such that the proportion, 
equation (1), must be modified by the introduction of a 
constant considerably greater than unity, into its ante- 
cedents. A motor of this kind, with few field turns com- 
pared to armature turns, might be suitable for traction, 
but more important improvements have been made, which 
will now be discussed. 



101. Compensated Series Motors. — From an inspection of 
Fig. 197 it is seen that the power factor of series motors may 
be increased by increasing IR and E rot , or by decreasing E f 
and E a . It is obvious that increasing IR signifies an 
increase in losses, thus resulting in a lower efficiency. 
E rot can be increased by increasing the number of armature 
turns. Both E f and E a can be decreased by lowering the 
frequency without affecting E roh hence low frequencies 
are desirable. To decrease the reactive electromotive 
force of the field, it is necessary that the reluctance 



MOTORS. 271 

of the magnetic circuit be low, i.e., small air gap and 
low flux densities in the iron, in order that the required 
flux can be produced by a minimum number of ampere 
turns. The armature reactive E.M.F., E a , is not essential 
to the operation of the motor, and can be neutralized by 
the use of compensating windings, and this feature of 
alternating- current series motors is a very important one. 

The compensating winding is embedded in slots in the 
pole faces, as shown in Fig. 198, which represents a West- 




Fig. 198= 

inghouse four-pole compensated single-phase railway motor 
with its armature and field windings removed. The num- 
ber of turns of the compensating winding is adjusted so 
as to set up a magnetomotive force equal and opposite to 
that due to the current in the armature coils. The com- 
pensating winding may be energized either by the main 
current, by placing this winding in series with field and 
armature, or by an induced current, which is obtained 
by short-circuiting the compensating winding upon itself, 
thus utilizing the principle of the transformer in that the 
main and induced currents are opposite in phase. The 
former method of neutralizing E a is known as conductive 
or forced compensation, and may be used with both alter- 



2*]2 



ALTERNATING-CURRENT MACHINES. 



nating and direct currents, and the latter method is known 
as inductive compensation, and may be used only with alter- 
nating current. 

Figs. 199 and 200 show schematically the connections of 





Fig. 199. 



Fig. 200. 



the conductively and inductively compensated alternating- 
current series motors respectively. The compensating 



Sxc/t/ng 




Co/7?pe/7<5a£/>?& 



winding is preferably distributed so that the armature 
reactance is neutralized as completely as possible. The 



MOTORS. 273 

current flows in the same direction in all of the con- 
ductors of the compensating winding embedded in one 
field pole, and flows in the opposite direction in the con- 
ductors embedded in the adjacent poles. Fig. 201 illus- 
trates a distributed conductive-compensating winding of a 
four-pole machine. 

When the compensating winding completely neutralizes 
the armature reactance, the impressed electromotive force 
(Eq. 7, §99) is 

E = V(E rot + IRf + e;, 

where R is the resistance of the motor including that of the 

compensating winding. If the resistance, R, be neglected, 

then, since 

V 

100 





DO/1 


:£f> 


the impressed elect 


romo 
= Ef 


tive force becomes 


E 


Aw, 


:h- 


and therefore the power 


factor is 




COS (f) 


E ro 
E 


t 


V 


~ VV 2 


+ (60 frf 


The motor current 


is 






I 




E 





WfekJ 



.60/1 

At synchronous speed V = 60/, and therefore the power 
factor at this speed becomes - — ^=^= • 

Vl + T 2 



274 ALTERNATING-CURRENT MACHINES. 

Still neglecting the motor resistance, the current intake 
at synchronous speed is ■ ■ . and at standstill it 

X f Vl + T 2 

is ■— - , consequently the ratio of the current at synchronous 
speed to the current at standstill is — — . Since 

Vl + T 2 

torque varies as the square of the current, the ratio of 
the torque at synchronous speed to the starting torque is 

T 2 

• Hence it follows that the speed-torque charac- 
teristics of a compensated series motor may be adjusted 
to the required conditions by properly proportioning the 
number of armature and field turns. 

Compensated series motors are well suited for traction. 
The performance curves of the 250-horse-power 25-cycle 
Westinghouse conductively compensated single-phase series 
motor used on the New York, New Haven and Hartford 
Railroad are shown in Fig. 202. Each locomotive is 
equipped with four of these motors operating at 225 volts, 
which is procured by step-down transformation from over- 
head 1 1, 000- volt trolley. On parts of the road the motors 
operate on direct current, the current being supplied directly 
to the motors. 



102. Sparking in Series Motors. — The principal diffi- 
culty encountered in the operation of single-phase series 
motors is the sparking at the brushes. This is caused by 
the local currents produced by the E.M.F. generated in 
the armature turns short-circuited by the brushes, due to 
the periodic reversals of the field flux. With the brushes 
located in the neutral position with respect to the E.M.F, 



POWER TRANSMISSION, 



275 



of rotation, the short-circuited turns are perpendicular to 
the axis of the field flux, and therefore the flux linked with 
these turns is a maximum. The electromotive force gener- 
ated in a short-circuited armature section is 



E s = 



2nf$ m N s 

V2 io 8 



where N s is the equivalent number of armature turns per 
section which is short-circuited by a brush. If the resist- 



500 100 



MILES PER HOUR 



100 









0^ 


^£2. F 


i££Oft 













V 




^f 


^cT 


^vcy 










\ 












/ 




























































K 


> 


















<<y 















































800 1000 1200 1400 1600 

AMPERES 



Fig. 202. 



ance of the section be small, an enormous current will flow, 
and will cause excessive heating of the brush, commutator 
segments, and armature conductors. 
In order to decrease this local current^ the E.M.F. 



276 ALTERNATING-CURRENT MACHINES. 

induced in each section may be decreased and the resist- 
ance thereof may be increased. From an inspection of 
the preceding formula it is seen that E s may be decreased 
by reducing the number of armature turns per section, by 
lowering the maximum value of the flux, and by lowering 
the frequency. Thus single-phase series motors are 
usually provided with more than two poles and with 
many commutator segments, and are designed to operate 
on low frequency circuits. A simple way to increase the 




Fig. 203. 

resistance of the armature sections involves the use of 
preventive or resistance leads, which are connected between 
armature conductors and commutator segments, as illus- 
trated in Fig. 203. It has been shown by experiment that 
the losses are a minimum when the resistance of the pre- 
ventive leads is so proportioned that the short-circuit cur- 
rents and normal currents are equal. The resistance 
leads are usually of German silver and have a large current- 
carrying capacity. They are placed in the same slots as 
the armature conductors, and usually at the bottom thereof. 
Only a few of these leads are in circuit at any instant, 
and, when the armature rotates, all of the preventive leads 
carry current in turn, hence the average loss of power per 



MOTORS. 



277 



lead is small. As the heating effect of the short-circuit 
current depends upon the duration of the short circuit, it 
is essential that the brushes be made quite narrow. 



103. Repulsion Motors. — The repulsion motor con- 
sists of a field resembling the stator of the single-phase 
induction motor, and an armature which is similar to the 
armatures of direct-current and alternating-current series 
motors. The armature winding always remains short- 




Fig. 204. 

circuited in a line inclined at a definite angle with the 
field axis, this being accomplished by means of brushes, 
bearing on the commutator, which are joined together 
by a conductor of low resistance. The field winding is 
supplied with single-phase alternating current. The fact 
that the armature and field windings are electrically dis- 
tinct makes it possible to operate the motor on high volt- 
age systems, the armature winding being so adjusted that 
the currents therein can be commutated satisfactorily. 



278 



ALTERNATING-CURRENT MACHINES. 



The pulsating flux through the armature, produced by 
the alternating current in the field winding, may be con- 
sidered as the resultant of two components, one in the 
direction of the brush axis, and the other perpendicular 
thereto; these being represented in Fig. 204 respectively 
by OA and AB. The component OA produces an E.M.F. 
in the armature conductors and causes a current to flow 




Fig. 205. 



through them. The other component, AB, reacts upon 
this armature current, thereby developing torque. 

To represent the action of a repulsion motor more 
clearly, the field winding may be considered as composed 
of two parts placed at right angles to each other, as at 
M and AT", Fig. 205, and the brushes may be located in 
line with one of them. When the rotor is stationary, the 
pulsating flux from poles MM causes the flow of current in 
the short-circuited armature, the effect of which is the pro- 
duction of a flux opposite and nearly equal to that which 
caused the current flow. The flux from poles MM is 
thereby reduced, thus resulting in increased stator current 



POWER TRANSMISSION. 



279 



and flux from poles AW. Neglecting iron losses, this flux 
will be in phase with the line current, whereas the phase of 
the armature current is opposite to that of the current in 
coils MM. Hence the flux from poles AW is in phase 
with the armature current, and their product, torque, 




Fig. 206. 

retains its sign as both reverse in direction together. The 
repulsion motor exerts its maximum torque at starting, and 
this torque decreases with decreasing current and with 
increased speed, and consequently this type of commuta- 
tor motor is well adapted for single-phase traction. The 
power factor of repulsion motors is low at starting and 
rises rapidly as the speed increases. Repulsion motors 
may be operated on 25 ~ or even on 60 ~ supply circuits. 
The fact that the repulsion motor may be converted 



280 



ALTERNATING-CURRENT MACHINES. 



readily into a single-phase induction motor, by simply 
short-circuiting the entire commutator, thus changing the 
armature to a squirrel-cage rotor, has led to the design of 
single-phase induction motors which start and come up 
to speed as repulsion motors. A motor of this type, manu- 

















. 


P£ 






































ED 




























































































































































































4 




kc 


* 




3 H.P. SINGLE PHASE 
INDUCTION'MOTOR 

FROM TESTS MADE AT 
HARVARD UNIVERSITY, 












tj 


& 












*#u 
































> 


r 














MAY 1900 










/ 




















140 VOLTS 












r 




















| 
















/ 


i 




















fcttfJ 
















7 


















0^ 




















/ 




H 




































f 


— 


i — 


















































































i 

































10 20 30 40 50 60 70 
% FULL LOAD 

Fig. 207. 



factured by the Wagner Electric Company, is shown in 
Fig. 206. Upon reaching normal speed, a centrifugal 
device, shown in the figure, causes the commutator bars 
to be short-circuited, and the brushes are simultaneously 
lifted from the commutator. The results of tests made 
upon this type of motor are represented in the curves of 
Fig. 207. 

104. Series-Repulsion Motor. — A single-phase railway 
motor which embodies many of the best features of the 
repulsion motor and of the compensated series motor, and 
therefore called the series-repulsion motor, has recently 
been developed by the General Electric Company. 



POWER TRANSMISSION. 



28l 



The windings resemble those of a series motor, and the 
armature has a fractional-pitch winding such as is used 
on direct-current motors. The connections of the motor 
circuits for the starting and running positions are shown 
in Fig. 208. 

The motor starts as a repulsion motor and possesses 
all the characteristics thereof when in this position. The 




— G 




— G 



Fig, 208. 



compensating coils differ from those on series motors in 
that they have twice as many turns as there are on the 
armature, and therefore the armature current at starting 
is twice as large as the current flowing through the field 
and compensating windings. Thus the starting torque is 
twice as great as would be obtained for the same current 
with a compensated series motor. When in the running 
position, the motor characteristics are similar to those of 
the compensated series motor, but it possesses an advantage 
over the latter in regard to sparking. 

At starting there is very little sparking at the brushes, 



282 



ALTERNATING-CURRENT MACHINES. 



but this increases up to a certain value of voltage induced 
in the short-circuited armature sections. This value 
practically corresponds to that which gives good com- 
mutation in the running position. Better inherent com- 
mutation is the chief advantage of the series-repulsion 
motor. 

PROBLEMS. 

i. Determine approximately the full-load efficiency of a certain 
15-horse-power three-phase six-pole 50-cycle induction motor which 
makes 950 revolutions per minute when carrying full load. 

2. What torque does the motor of the preceding problem exert 
when operating under full load ? 

3. Calculate the leakage reactance per phase of a 5000-volt three- 
phase 25-cycle 30-pole induction motor having a three-phase wound 




Fig. 209. 

rotor (both stator and rotor have Y-connected full-pitch windings). 
The dimensions of the slots in inches are indicated in Fig. 209, and 
the constants of the motor are: 

Rotor diameter. 118 in. 

Total number of primary slots 450 

Total number of secondary slots. . . 720 

Series conductors per primary slot 8 

Series conductors per secondary slot .... ....... § 

Length of slots. 29.5 in. 

Length of end connection per primary turn 48.5 in. 

Depth of laminations back of slots 2.36 in. 



PROBLEMS. 283 

4. Calculate the exciting current per phase of the induction motor 
of the preceding problem. 

5. The stator resistance per phase of the induction motor of prob- 
lem 3 is 1,7 ohms, and the rotor resistance per phase is .02 ohm, 
which, when reduced to the stator circuit, is 2 ohms. Determine the 
motor performance curves from its circle diagram, this being based 
upon the results of the two preceding problems. 

6. Plot curves of P x and P 2 for a single-phase synchronous motor, 
excited so as to generate 1000 volts, connected to a single-phase alter- 
nator having an E.M.F. of 1200 volts, the total resistance of the 
circuit being 1.75 ohms and the total impedance 2 ohms. 



284 



ALTERNATING CURRENT MACHINES. 



CHAPTER VIII. 



CONVERTERS. 



105. The Converter. — The converter is a machine hav- 
ing one field, and one armature, the latter being supplied 
with both a direct-current commutator and alternating- 
current slip-rings. When brushes, which rub upon the 
slip-rings, are connected with a source of alternating 
current of proper voltage, the armature will rotate syn- 
chronously, acting the 
same as the armature of a 
synchronous motor. While 
so revolving, direct current 
can be taken from brushes 
rubbing upon the commu- 
tator. The intake of cur- 
rent from the alternating- 
current mains is sufficient 
to supply the direct-current 
circuit, and to overcome 
the losses due to resistance, 
friction, windage, hyster- Flg * 2I °- 

esis, and eddy currents. The windings of a converter 
armature are closed, and simply those of a direct-current 
dynamo armature with properly located taps leading to the 
slip-rings. Each ring must be connected to the armature 
winding by as many taps as there are pairs of poles in 
the field. These taps are equidistant from each other. 




CONVERTERS. 



285 



There may be any number of. rings greater than one. 
A converter having 11 rings is called an //-ring converter. 

The taps to successive rings are -th of the distance be- 

11 

tween the centers of two successive north poles from each 
other. Fig. 210 shows the points of tapping for a 3-ring 
multipolar converter. 

A converter may also be supplied with direct current 




Fig. 211. 

through its commutator, while alternating current is taken 
from the slip-rings. Under these circumstances the 
machine is termed an inverted converter. Converters are 
much used in lighting and in power plants, sometimes 
receiving alternating current, and at other times direct 
current. In large city distributing systems they are often 
used in connection with storage batteries to charge them 



286 ALTERNATING-CURRENT MACHINES. 

from alternating-current mains during periods of light 
load, and to give back the energy during the heavy load. 
They are also used in transforming alternating into direct 
currents for electrolytic purposes. A three-phase machine 
for this purpose is shown in Fig. 211. 

A converter is sometimes called a rotary converter or 
simply a rotary. 

106. E.M.F. Relations. — In order to determine the re- 
lations which exist between the pressures available at the 
various brushes of a converter, 

Let E d = the voltage between successive direct-current 
brushes. 
E n — the effective voltage between successive rings 
of an n-r'mg converter. 

a = the maximum E.M.F. in volts generated in a 
single armature inductor. This will exist 
when the conductor is under the center of a 
pole. 

b = the number of armature inductors in a unit 
electrical angle of the periphery. The 
electrical angle subtended by the centers of 
two successive poles of the same polarity 
is considered as 2ir 

The E.M.F. generated in a conductor may be considered 
as varying as the cosine of the angle of its position relative 
to a point directly under the center of any north pole, the 
angles being measured in electrical degrees. At an angle 
/?, Fig. 212, the E.M.F. generated in a single inductor G 
is a cos ft volts. In an element d(3 of the periphery of 
the armature there are bdfi inductors, each with this 
E.M.F. If connected in series they will yield an E.M.F. 



CONVERTERS. 



287 



of ab cos fi dfi volts. The value of ab can be determined 
if an expression for the E.M.F. between two successive 

direct-current brushes be 
P^X determined by integration, 
and be set equal to this 
value E d as follows : 







<=/: 



E fl = 



ab cos /3d/S = 2 ab. 



ab 



Fig. 212. 



successive rings is 



2tt 



In an /z-ring converter, the 
electrical angular distance 
between the taps for two 

The maximum E.M.F. will be 



generated in the coils between the two taps for the succes- 
sive rings, when the taps are at an equal angular distance 
from the center of a pole, one on each side of it, as shown 
in the figure. This maximum E.M.F. is 






ab cos j3dft = 2 ab sin 



= E d sin - . 
n 



The effective voltage between the successive rings is 
therefore 

77 E d . 7T 

By substituting numerical values in this formula, it is 
found that the coefficient by which the voltage between 



288 



ALTERNATING-CURRENT MACHINES. 



the direct-current brushes must be multiplied in order to 
get the effective voltage between successive rings is for 

2 rings 0.707 

3 rings 0.612 

4 rings 0.500 

6 rings 0.354 

In practice there is a slight variation from these co-effi- 
cients due to the fact that the air-gap flux is not sinusoid- 
ally distributed. 

107. Current Relations. — In the following discussion it is 
assumed that a converter has its field excited so as to 
cause the alternating currents in the armature inductors to 
lag 180 behind the alternating E.M.F. generated in them. 

The armature coils carry currents which vary cyclically 
with the same frequency as that of the alternating-current 
supply. They differ 
widely in wave-form from 
sine curves. This is be- 
cause they consist of two 
currents superposed upon 
each other. Consider a 
coil B, Fig. 213. It car- 
ries a direct current whose 

value — is half that car- 
2 

ried by one direct-current 

brush, and it reverses its 

direction every time that Fl£ * 2I3 ' 

the coil passes under a brush. The coil, as well as 

all others between two taps for successive slip-rings, also 

carries an alternating current. This current has its zero 




CONVERTERS. 



289 



value when the point A, which is midway between the 
successive taps, passes under the brush. The coil being 
<[> electrical degrees ahead of the point A, the alternating 

current will pass through zero -*— of a cycle later than 

2 7Z 

the direct current. The time relations of the two currents 
are shown in Fig. 214. 

To determine the maximum value of the alternating 
current consider that, after subtracting the machine losses, 




Fig. 214. 

the alternating-current power intake is equal to the direct- 
current power output. Neglecting these losses for the 
present, if E n represents the pressure and I n the effective 
alternating current in the armature coils between the suc- 
cessive slip-rings, then for the parts of the armature wind- 
ings covered by each pair of poles 

E d I d = nEJ n 



V 



sin -I n . 



Therefore, the maximum value of the alternating current is 



2 h = 



. 71 

wsin- 
n 



The time variation of current in the particular coil B is 
obtained by taking the algebraic sum of the ordinates of 



290 



ALTERNATING-CURRENT MACHINES. 



the two curves. This yields the curve shown in Fig. 215. 
Each inductor has its own wave-shape of current, 
depending upon its angular distance (p from the point A. 
Converter coils, therefore, alternately functionate as motor 
and as generator coils. 




Fig. 215. 

108. Heating of the Armature Coils. — The heating 
effect in an armature coil due to a current of such peculiar 
wave-shape as that shown in Fig. 215 can be determined 
either graphically or analytically. The graphic determina- 
tion requires that a new curve be plotted, whose ordinates 
shall be equal to the squares of the corresponding current 
values. The area contained between this new curve and 
the time axis is then determined by means of a planimeter. 
The area of one lobe is proportional to the heating value 
of the current. This value may be determined for each 
of the coils between two successive taps. An average of 
these values will give the average heating effect of the 
currents in all the armature coils. The heating is different 
in the different coils. It is a maximum for coils at the 
points of tap to the slip-rings and is a minimum for coils 
midway between the taps. 



CONVERTERS. 291 

109. Capacity of a Converter. — As the result of a 
rather involved analysis it is found that a machine has 
different capacities, based upon the same temperature rise, 
according to the number of slip-rings, as shown in the fol- 
lowing table. The armature is supposed to have a closed- 
coil winding. 

CONVERTER CAPACITIES. 
Used as a Kilowatt Capacity 

Direct-current generator . c 100 

Single-phase converter 85 

Three-phase converter 134 

Four-phase converter 164 

Six-phase converter 196 

Twelve-phase converter 227 

The overload capacity of a converter is limited by com- 
mutator performance and not by heating. As there is but 
small armature reaction, the limit is much higher than is 
the case with a direct-current generator. 

no. Starting a Converter. — Converters may be started 
and be brought up to synchronism by the same methods 
which are employed in the case of synchronous motors. 
It is preferable, however, that they be started from the 
direct-current side by the use of storage batteries or other 
sources of direct current. They may be brought to a little 
above synchronous speed by means of a starting resistance 
as in the case of a direct-current shunt motor, and then, 
after disconnecting and after opening the field circuit, the 
connections with the alternating-current mains may be 
made. This will bring it into step. 

in. Armature Reaction. — The converter armature cur- 
rents give rise to reactions which consist of direct-current 



292 



ALTERNATING-CURRENT MACHINES. 



generator armature reactions superposed upon synchronous 
motor armature reactions. It proves best in practice to 
set the direct-current brushes so as to commutate the cur- 
rent in coils when they are midway between two succes- 




Fig. 216. 

sive poles. The direct-current armature reaction, then, con- 
sists in a cross-magnetization which tends to twist the field 
flux in the direction of rotation. When the alternating 
currents are in phase with the impressed E.M.F. they also 
exert a cross-magnetizing effect which tends to twist the 



CONVERTERS. 293 

field flux in the opposite direction. The result of this neu- 
tralization is a fairly constant distribution of flux at all 
loads. Within limits even an unbalanced polyphase con- 
verter operates satisfactorily. There is no change of field 
excitation necessary with changes of load. 

The converter is subject to hunting the same as the 
synchronous motor. As its speed oscillates above and 
below synchronism, the phase of the armature current, in 
reference to the impressed E.M.F., changes. This results 
in a distortion of the field flux, of varying magnitude. 
This hunting is much reduced by placing heavy copper 
circuits near the pole horns so as to be cut by the oscillat- 
ing flux from the two horns of the pole. The shifting of 
flux induces heavy currents in these circuits which oppose 
the shifting. Fig. 216 shows copper bridges placed be- 
tween the poles of a converter for this purpose. 

When running as an inverted converter from a direct- 
current circuit, anything which tends to cause a lag of the 
alternating current behind its E.M.F. is to be avoided. 
The demagnetization of the field by the lagging current 
causes the armature to race the same as in the case of an 
unloaded shunt motor with weakened fields. Converters 
have been raced to destruction because of the enormous 
lagging currents due to a short circuit on the alternating- 
current system. 

112. Regulation of Converters The field current of a 

converter is generally taken from the direct-current 
brushes. By varying this current the power factor of the 
alternating-current system may be changed. This may 
vary, through a limited range, the voltage impressed 
between the slip-rings. As the direct-current voltage 



2 9 4 



ALTERNATING-CURRENT MACHINES. 



Step-down 

Transformer. "«*"'*<* 



bears to the latter a constant ratio it may also be varied. 
This is, however, an uneconomical method of regulation. 
Converters are usually fed through step-down transform- 
ers. In such cases 
there are two com- 
mon methods of regu- 
lation, which vary the 
voltage supplied to 
Fig - 2I7 ' the converter's slip- 

rings. The first is the method of Stillwell, which is 
shown in the diagram, Fig. 217. 

The regulator consists of a transformer with a sectional 





Fig. 518. 



CONVERTERS. 295 

secondary. Its ratio of transformation can be altered by 
moving a contact-arm over blocks connected with the 
various sections, as shown in the diagram. The primary 
of the regulator is connected with the secondary terminals 
of the step-down transformer. The sections of the second- 
ary, which are in use, are connected in series with the step- 
down secondary and the converter windings. 

The second method of regulation is that employed by 
the General Electric Co. The ratio of transformation of 
a regulating transformer, which is connected in circuit in 
the same manner as the Still well regulator, is altered by 
shifting the axes of the primary and secondary coils in 
respect to each other. Fig. 218 shows such a transformer, 
the shifting being accomplished by means of a small, 
direct-current motor mounted upon the regulator. The 
primary windings are placed in slots on the interior of a 
laminated iron frame, which has the appearance of the 
stator of an induction motor. The secondary windings are 
placed in what corresponds to the slots of the rotor core. 
The winding is polar ; and if the secondary core be rotated 
by an angle corresponding to the distance between two 
successive poles, the action of the regulator will change 
from that of booster to that of crusher. 

Another method of converter regulation, sometimes 
used in railway work, makes use of reactance coils, con- 
nected between the step-down transformer coil terminals 
and the slip-rings of the converter, as well as of an ordi- 
nary series compounding coil on the field-cores of the con- 
verter. The series and shunt field coils are so adjusted 
that the converter takes a lagging current at no load and 
a leading current at full load. The step-down transformer 
voltage being assumed as constant, the voltage impressed 



296 ALTERNATING-CURRENT MACHINES. 

upon the slip-rings will be the remainder resulting from 
the vector subtraction of the reactance drop from the 
constant voltage. On heavy loads and leading currents 
this remainder is greater than the constant voltage. There 
is therefore a constantly increasing voltage impressed 
upon the slip-rings as the load increases. Too large a 
reactance, however, is liable to introduce pulsation troubles. 

In Europe some use is made of a small auxiliary alter- 
nator mounted upon the shaft of the converter and oper- 
ating synchronously with it. By varying and reversing 
the field excitation of this alternator, whose armature 
phases are connected between the transformer terminals 
and the slip-rings of the converter, it may be caused to 
act as a booster or as a crusher. 

Recently converters have been constructed in a manner 
that permits of altering their ratios of voltage conversion 
by changing the distribution of flux in the air gap. Non- 
sine waves of E.M.F. are then induced in the armature 
inductors. The ratios of voltage conversion hitherto 
deduced upon the assumption of sine wave-forms do not 
then hold. The change of flux distribution is accomplished 
by splitting each pole into sections along axial planes. 
The sections are then subjected to different magneto- 
motive forces which may be independently varied during 
operation. 

113. Mercury Vapor Converter. — A mercury vapor 
converter, which is suitable for use in charging storage 
batteries from a single-phase circuit, is shown with its con- 
nections in Fig. 219. It consists of a very highly exhausted 
glass bulb equipped with four electrodes, of which two 
are positive, one negative, and the other an auxiliary 
which is used only in starting. The two latter electrodes 



CONVERTERS. 



297 



are of mercury. The two external terminals of an auto- 
transformer are connected with the two positive electrodes, 
while the internal terminals are connected to the single- 
phase supply circuit. The operation of this converter is 
based upon the facts that 
(a) to start a current be- 
tween two electrodes in 
a vacuum bulb of this 
character there must be 
impressed upon these 
electrodes a very high 
voltage (25, 000 volts), most 
of which is consumed in 
overcoming a transition 
resistance at the negative 
electrode, and (b) once 
started this cathode tran- 
sition voltage drops to a 
very small value (4 volts). 
In operation, and after 
starting, therefore, current 
flows during one-half of a 
cycle from the left-hand 
terminal of the trans- 
former to the left-hand 
positive electrode through 
the vapor to the main 
negative electrode and thence through the battery to the 
center of the transformer coil, and during the following 
half cycle flows from the right-hand terminal to the right- 
hand positive electrode through the tube and battery as 
before. The positive electrodes permit current to flow 




Fig. 219. 



298 ALTERNATING-CURRENT MACHINES. 

from them into the tube but never in the reverse direction. 
They are therefore each idle during alternate half cycles. 
The transition resistance of the negative main electrode 
when once broken down remains so as long as current 
enters it from the vapor. 

To start the converter the bulb is tilted until there is 
a mercury connection between the main negative and 
auxiliary electrodes. This permits a current to flow from 
the storage battery through the mercury into the auxiliary 
electrode. If now the mercury bridge be broken, by 
restoring the bulb to its original position, vapor conduction 
will be established between the main negative and aux- 
iliary electrodes. The transition resistance of the latter 
is thus broken down and the converter begins to operate, 
current flowing alternately from the two positive electrodes 
to the auxiliary electrode. If now the converter be again 
tilted and restored to its normal position the point of 
entrance of the vapor current into the mercury can be 
transferred to the main negative electrode. A second 
tilting is seldom necessary, the mercury generally making 
several makes and breaks of the circuit during the first 
tilt as a result of its fluidity. If for an instant (one mil- 
lionth of a second) the current ceases to enter the mercury, 
the cathode transition resistance will reestablish itself. 
An inductance inserted in the battery circuit causes a 
sufficient lag of current behind the voltage between a 
positive and the negative electrode to enable the voltage 
due to the other positive electrode to maintain the opera- 
tion. The current in the battery circuit is unidirectional 
but pulsating. 



PROBLEMS. 299 



PROBLEMS. 

1. From what points on the armature winding should taps be taken 
for connection with the successive rings of a 5-ring 6-pole converter? 

2. A 4-pole converter is supplied with six slip-rings so as to be 
adapted for use on single-, two-, or three-phase circuits. The rings 
used on single-phase are 1 and 4; on two-phase are 1 and 4, and 2 and 
6; on three-phase 1, 3, and 5. Locate the points of attachment of taps 
from each ring to the armature winding. 

3. A 12-ring converter delivers 600 volts to a direct-current railway 
circuit. What is the voltage between successive slip-rings? 

4. A 20-pole 6-ring converter delivers 1000 amperes of direct current 
at full load. Neglecting armature resistance and other losses, determine 
the current wave-shape in a conductor 20 electrical degrees in advance 
of tap to a slip-ring. 

5. During -what portion of a revolution is the current in the con- 
ductor mentioned in problem 4 so directed as to exert a motor effort ? 

6. Compare the heating effect of full-load current in the conductor 
of problem 4 with that in a conductor midway between taps. 



300 ALTERNATING-CURRENT MACHINES. 



CHAPTER IX. 

POWER TRANSMISSION. 

114. Superiority of Alternating Currents. — In trans- 
mitting power electrically over long distances, it is neces- 
sary to employ high voltages, so that, with a reasonable 
line loss, the cost of the conductors will not be excessive. 
In the United States, power transmission at high voltages 
has been accomplished by means of alternating current 
only. In Europe, however, considerable attention has 
been given to the development of the Thury system of 
direct-current transmission. There are a number of 
plants successfully employing this system at the present 
time, but the highest voltages used are in the neighbor- 
hood of 20,000, and the amounts of power transmitted are 
comparatively small. If the line alone be considered, 
direct current is far superior to alternating current. The 
former has unity power factor, is free from inductive dis- 
turbances, such as surges, and it has no wattless charging 
current to reduce the effective output of the machines. 
As will be shown later, the amount of conductor material 
required in a direct-current line is less than that required 
in an alternating-current line with the same maximum 
voltage in the two cases. 

A comparison of the station apparatus of both systems 
of power transmission shows that the direct-current system 
is at a great disadvantage. Three thousand volts is the 



POWER TRANSMISSION. 301 

maximum that can be successfully handled on a com- 
mutator, even with the special design of machine such 
as Thury has developed. Consequently, to obtain the 
required high line voltage, a number of generators must be 
connected in series, series-wound machines being used. 
When a machine is generating 3000 volts, the maximum 
current that can be commutated is about 100 amperes, so 
that the individual machines have small output. Each 
generator must be insulated from ground, and, as several 
machines are connected to the same prime mover, they 
must be insulated therefrom and from each other. The 
system is grounded at the middle point, so as to limit the 
amount of insulation required; that is, the insulation 
under each machine must be capable of withstanding the 
maximum difference of potential between its terminals 
and ground. The line current is maintained constant by 
several complicated auxiliary devices. These automatic- 
ally regulate the speed of all the prime movers, so as to 
keep the line voltage proportional to the load; cut in or 
out of circuit one or more machines if there be large 
changes in load, and short-circuit any disabled machine. 

In the substations a number of series-wound motors 
are connected in series across the line, the motors being 
arranged in groups, each group driving a generator. The 
generators, which may deliver either direct or alternating 
current, are connected in multiple for distribution. The 
motors and generators in the substation must be insulated 
from each other and from ground, just as are the machines 
in the generating station. The current taken by the 
motors is kept constant by an automatic shifting of the 
brushes. The Thury system is adapted only to under- 
takings where the power is to be transmitted over a long 



302 ALTERNATING-CURRENT MACHINES 

distance and the load is to be concentrated at few points^ 
since at every tap a complete substation must be provided 
containing motors having an aggregate voltage equal to 
the line voltage. On the other hand, in an alternating- 
current system, a static transformer can be installed any- 
where along the line and it will operate satisfactorily with 
practically no attention. 

Considering the line, alone, the employment of direct 
current is better and more economical than that of alter- 
nating current. But when the whole plant, including 
generating station, line, and substations, is considered, 
the employment of the alternating-current system is held 
by many engineers to be the most advantageous. The 
alternating-current system is more reliable, more flexible, 
and, with the exception of special cases, is probably 
cheaper than the direct-current system, in spite of the 
greater cost of the line conductors. 

115. Frequency. — According to the Standardization 
Rules of the A. I. E. E., there are two standard fre- 
quencies, namely, 60 cycles and 25 cycles. In early 
transmission plants the frequency employed was 60 cycles 
or higher. All recent transmissions, however, are at 25 
cycles, and there is a strong tendency to lower this 
frequency to 15 or even to 12^ for certain classes of 
work. Sixty-cycle generators and transformers are smaller 
and cheaper than are those of lower frequency. It was 
formerly thought that for lighting, a frequency higher than 
25 cycles was necessary in order to prevent flickering of 
the lamps. But the success of 25-cycle lighting in Buffalo 
from the circuits of the Niagara Falls Power Company 
has proved that, if the form factor of the voltage wave is 



POWER TRANSMISSION. 303 

not greater than that for a sine wave, the higher frequency 
is unnecessary. The Niagara generators give a wave 
slightly flatter than a sine wave; and all modern genera- 
tors of large output can be depended upon to give good 
wave forms. 

The advantages of low frequency for transmission lines 
are as follows: (a) The inductive drop, 2 nfLI, is less, and 
consequently the regulation is better than for high fre- 
quencies. 

(b) The capacity current, 2 nfEC, also increases with 
the frequency. Its effect is to reduce the energy output of 
the generators and transformers. 

(c) The lower the frequency, the less difficult becomes 
the problem of operating generators and other synchronous 
apparatus in parallel. This is because the unavoidable 
variations in speed are smaller in proportion to the angular 
velocity, the lower the frequency. 

(d) The power factor of an induction motor decreases 
as the frequency is raised. This is an extremely important 
reason for using a low frequency, since the power load 
generally constitutes a large part of the total load of a 
transmission system. 

(e) A low frequency is also less liable to set up elec- 
trical oscillations as a result of the coincidence of the 
natural frequency of the line with that of an odd har- 
monic of the impressed E.M.F. If the distributed induc- 
tance and capacity of the line be L henrys and C farads 
respectively, then its natural frequency, as shown by 
Steinmetz, is to be expressed as 



aVLC 



304 ALTERNATING-CURRENT MACHINES. 

If the resistance be sufficiently low, as is often the case, 
oscillations at this frequency are liable to occur. A triple 
harmonic of some magnitude usually exists in the E.M.F. 
wave of each phase winding of an alternator. This does 
not appear at the terminals of a three-phase machine 
whether Y- or A-connected. It does appear, however, 
between the terminals and a grounded neutral. In the 
armature windings there is usually a triple harmonic com- 
ponent of current which sets up an armature reaction 
causing magnetic field distortion that results in fifth and 
seventh harmonic E.M.F's. Triple harmonics of E.M.F. 
or of current also result from the use of transformers. In 
three-phase work their influence upon the line may be 
overcome by the use of A connections. 

With lines constructed in accordance with present 
practice, the natural frequency for a length of 150 miles is 
about three hundred. This is the same as that of the 
fifth harmonic on a 60-cycle system, whereas for a fre- 
quency of 25 the fifth harmonic frequency would be but 
125. It would therefore be unwise to select a frequency of 
60 cycles for such a line. 

116. Number of Phases. — A comparison of the weights 
of line wire of a given material, necessary to be used in 
transmitting a given power, at a given loss, over the same 
distance, must be based upon equal maximum voltages 
between the wires. For the losses by leakage, the thick- 
ness and cost of insulation, and perhaps the risk of danger 
to life, are dependent upon the maximum value. A com- 
parison upon this basis gives, according to Steinmetz, the 
following results : — 



POWER TRANSMISSION. 305 

Relative weights of line wire to transmit equal power over 
the same distance at the same loss, with unit power factor. 

2 Wires. Single-phase 100. o 

Continuous current 5°-° 

3 Wires. Three-phase 75 .0 

Quarter-phase 145 . 7 

4 Wires. Quarter-phase 100. o 

The continuous current does not receive the approval 
of American engineers, as previously stated. The single- 
phase and four-wire quarter-phase system each requires 
one-third more wire than the three-phase system. 

By use of the Scott three-phase quarter-phase trans- 
former, the transmission system may be three-phase, 
while the distribution and utilization system may be 
quarter-phase. 

Each conductor of a three-phase line must be of the size 
required in a single-phase line transmitting half as much 
power, with the same percentage of loss, at the same volt- 
age and distance between conductors. 

117. Voltage. — If the frequency, the amount of trans- 
mitted power, and the percentage of power lost in the line, 
remain constant, the weight of line wire will vary inversely 
as the square of the voltage impressed upon the line. 
This depends upon the fact that the cross-section of the 
wire is not determined by the current density and the limit 
of temperature elevation, but by the permissible voltage 
drop. If the impressed voltage on a line be multiplied by 
n, the drop in the line may be increased n times without 
altering the line loss. For the line loss is to the total 
power given to the line as the drop in volts is to the 



306 ALTERNATING-CURRENT MACHINES, 

impressed voltage. To transmit the same power, but - th 

TV 

the previous current is necessary; and this current, to pro- 
duce n times the drop, must, therefore, traverse a resistance 
n 2 times as great as previously. 

In a long transmission line the conductors constitute 
one of the largest items, if not the largest item, of invest- 
ment of the entire plant. Consequently it is desirable to 
have the voltage as high as possible. But raising the 
voltage increases the investment for transformers, switch- 
ing apparatus, lightning protection, and insulators; and the 
depreciation and repair charges on these items are much 
greater than the corresponding charges on the conductors. 
The economic voltage to be employed for transmitting a 
given amount of power over a certain distance is that 
voltage which will lead to the minimum annual cost for 
the entire plant. Theoretically, this economic voltage can 
be determined by expressing the several elements of cost 
as functions of the voltage and equating the differential 
of this expression to zero. This method is complicated, 
and it requires so many assumptions as to render it of 
little use. 

118. Economic Drop. — A more practical way of deter- 
mining the voltage is based upon the fact that there are 
certain standard voltages for high-tension transformers. 
Except for special cases, a standard voltage should be 
used. For a given voltage, the amount of conductor 
material varies inversely as the drop, whereas the line loss 
varies directly with the drop. If the economic drop, 
which fixes the cross-section of the conductor, be calcu- 
lated for the several standard voltages, the best voltage to 



POWER TRANSMISSION. 307 

employ can readily be determined. For a given voltage 
at the generating station, the economic drop and cross- 
section of conductor for a single-phase circuit may be 
found as follows: 

Let E = voltage at generating station, 

P = power in kilowatts at generating station, 

L x = length of line in miles, i.e., length of a single 

conductor, 
R = total resistance of line in ohms, 
x = loss in terms of impressed quantities, 
S = section of conductor in circular mils, 
c t = cost of energy in dollars per kilowatt-year at 

generating station, 
c 2 = cost of conductor in dollars per pound, 
p 2 = interest rate on cost of line conductors, 
K x = resistance in ohms per mile of conductor hav- 
ing one circular mil cross-section, and 
K 2 = weight in pounds per mile of conductor having 
one circular mil cross-section. 

Then, line loss = Px. 

Annual cost of line loss = c x Px. 
Weight of line conductors = 2 K 2 L t S. 
Cost of line conductors = 2 c 2 K 2 L x S. 
Annual cost of line conductors = 2 p 2 c 2 K 2 L l S. 

2 L 
Line resistance = R = K t — — l • 

x . , ^ 1000 P „ 2000 PK X L< 

Line drop = Ex = - — - — R = — • 

E ES 

Section of conductor = S = - — ■=—— K< • 

E 2 x 



308 ALTERNATING-CURRENT MACHINES. 

The total annual charge due to line loss plus interest on 
conductors is 

c x Px + 2 p 2 c 2 K 2 L l S i 

and per delivered kilowatt is 

_ c x Px + 2 p 2 c 2 K 2 L x S 
q ~ ~ P - Px 

Substituting the value of S, this becomes 



(i) 



o , a t- t IOO ° P v 2 A 
^P.T + 2 p^C 2 K^L l — — — A x 

E- x 

P (i - x) 



(2) 



c.x ; p 2 c^K,Ko 4 A 2 i,ooo , x 

or ? = — ! — + y2 - \ y (3) 

1 - x E 2 x (1 - x) 



If K is substituted for p 2 c 2 K l K 2 4 Z 2 2 1000, then 

c,* K 
q = — ! — - + • 

1 — x E 2 x (1 — x) 



(4) 



To find the minimum value of q, its derivative is placed 
equal to zero, and there obtains 

dq . , 2 K K , N 

^ = ^+— .t-- = o; ( 5 ) 

whence * _ _ JL ± [(JLJ+ JL]\ 

But as x is positive 

x —^ + -k VKt + E ^ K - (6) 

If the preceding were worked out for a constant or fixed 
delivered E.M.F., instead of a fixed impressed E.M.F., 
allowing the latter to become what it might, the expression 



POWER TRANSMISSION. 309 

for q would be the same, except that the denominator 
would be P instead of P (1 — x), the quantities E, x, P, 
etc., being then delivered quantities instead of impressed 
quantities. If this be done, and the value of q be differ- 
entiated, there obtains 



Hence 



dq _ K 

dx~~ Cl ~ E 2 x 2 ~ °" 



— - = c t x. that is, the well-known relation 
E~x 



Interest = Loss. 

A three-phase line requires three-quarters as much con- 
ductor material as a single-phase line transmitting the 
same amount of power with the same loss. Each con- 
ductor of a three-phase transmission line has one-half the 
area of each conductor of the equivalent single-phase line. 
To find the economic drop for a three-phase line, multiply 
2 p 2 c.,K 2 L l S in equation (1) by f. Then f K will appear 
in equation (4) instead of K. Solving for the economic 
drop, 

x = - 3. . JL + -L^ V 9 K> + 12 E' Cl K. (7) 

4 E c x 4 E c t 

The area of each conductor is 

119. Line Resistance. — The resistance of anything but 
very large lines is the same for alternating currents as for 
direct currents. In the larger sizes, however, the resist- 
ance is greater for the alternating currents. The reason 
for the increase is the fact that the current density is not 



3io 



ALTERNATING-CURRENT MACHINES. 



uniform throughout a cross-section of the conductor, but 
is greater toward its outside. The lack of uniformity of 
density is due to counter electromotive forces set up, in 



T 

20 

Id 

o 

z 

£ 16 










































1- 1° 
tn 

UJZ, 
C£ u 

Ouj 

w 

as ? 

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10 20 30 40 50 60 70 

MILLIONS 
CIRCULAR MILS X FREQUENCY 

Fig. 220. 



90 



100 



the interior of the wire, by the varying flux around the 
axis of the wire which accompanies the alternations of the 
current. This phenomenon is termed skin effect, §21. Its 
magnitude may be determined from the curve, Fig. 220. 



120. Line Inductance. — The varying flux which is set 
up between the two line wires of a single-phase trans- 
mission circuit by the current flowing in them gives rise 
to a self-induced counter E.M.F. The inductance per 
unit length of single wire is numerically equal to the flux 
per unit current, which links a unit length of the line. 
To determine this value consider a single-phase line, with 
wires of R cms. radius, strung with d cms. between their 
centers, and carrying a current i. Let a cross-section of 



POWER TRANSMISSION. 



3U 



the line be represented in Fig. 221. The flux d& l} which 
passes through an element dr wide and of unit length, is 



& 




Fig. 221. 

equal to the magnetomotive force divided by the reluc- 
tance, or 

«**, = ±2 . 

2 Tir 
dr 

Integrating for values of r between d ~ R and R, 

'd - R s 



<!>! = 2 i log 



R 



and practically = 2 i log ( - J . 

There is some flux which surrounds 
the axis of the right-hand wire, and 
which lies inside the metal. This is of 
appreciable magnitude owing to the 
greater flux density near the wire. 
Represent the wire by the circle in 
Fig. 222, and suppose that the current 
is uniformly distributed over the wire. Then the current 




Fig. 222. 



312 ALTERNATING-CURRENT MACHINES. 



inside the circle of radius x is — i, and the magneto- 

R 

motive force which it produces is 

x 2 . 

A 7Z I. 

4 ^ 2 

The flux, however, which it produces links itself with 
but 7^-ths of the wire. The flux through the element dx, 

which can be considered as linking the circuit, is therefore 
, - 2 xH dx 

d ^ -&- ■ 

Integrating for values of x between o and R y 

^ 2 i 

For copper or aluminum wires /* = i. Hence the total 
flux linked with the line is 

* 1 + *^4iog,(4) + j], 

and the inductance, in absolute units, being the flux per 
unit current, is 

'-■*©+;• 

This gives by reduction the inductance in henrys per wire 
per mile as 

'd 



L = 80.5 + 740 log 



10 



The drop in volts due to the inductance, per mile of line 
(two conductors) per unit current, is therefore 



POWER TRANSMISSION. 313 

E L = .00405 / 2.3 log (-\ + -\ 

where d and R must be in terms of the same unit. 

It will be noted that the inductance depends upon 
the distance between conductors. This distance should 
increase with the voltage, but there is no definite relation 
between them. The following values represent average 
practice for bare overhead conductors: 



Kilovolts. 


Distance between Con- 
ductors in Inches. 


2.3 to 6.6 
10 to 20 
20 to 30 
3° to 50 
50 to 60 


28 
40 
48 
60 

72 



121. Line Capacity. — The two conductors of a single- 
phase transmission line, together with the air between 
them, act as a condenser. The conductors correspond to 
the condenser plates, and the air corresponds to the dielec- 
tric. When the lines are long, or when the conductors are 
close together, the capacity is quite appreciable. 

The capacity between two parallel cylindrical con- 
ductors may be determined as follows: Let A and B 
(Fig. 223) represent the two conductors of R cms. radius 
and d cms. apart between centers. Let A be charged with 
+ Q electrostatic units of electricity per centimeter of length, 
and B with — Q units. If the charge on A be alone con- 
sidered, there emanates from each unit length an electro- 
static flux of 4 nQ lines directed radially away from the axis 
of A. Similarly, a flux of — 4 nQ emanates radially from 
each unit length of B, due to its charge. The negative sign 



314 



ALTERNATING-CURRENT MACHINES. 



indicates that the flux is directed towards the axis of B. 
The superposition of these two fluxes results in an electro- 
static field such as would exist if a neutral conducting 




Fig. 223. 

plane were introduced halfway between A and B and 
perpendicular to the plane of their axes, and the potential 
of the neutral conducting plane were maintained as much 
below that of A as it is above that of B. To determine 
the potential differences, consider that the difference of 
potential between two points is equal to the work that 
must be performed on a unit positive charge to move it 
from one point to the other. The intensity of the field at a 
point C, at a distance x from the axis of A, due to the 
charge on A, this intensity being the flux-density or force 
which would be exerted upon a unit positive charge, is 

4 7ZQ 2 Q _ 

2 7ZX X 

and that due to the charge on B, noting that it is in the 
same direction as that due to A, is 



Fb = 



2 X (d — x) d — x 



4 *<2 



POWER TRANSMISSION. 315 

Therefore the total intensity or force is 

Hence the difference of potential between A and the 
neutral plane, which is the same as that between the 
neutral plane and B, is 

The capacity between either conductor and the neutral 
plane is therefore 

r Q * 

G ~ E , T^R.' 

slog,^- 

in electrostatic units per centimeter length of conductor. 
In transmission lines, R is usually so small compared with 
d as to be neglected. Reducing to miles, microfarads, and 
common logarithms, the capacity between neutral plane 
and either conductor is 

C = -^3 — _ microfarads per mile, 



'O 



and between conductors it is half as much. 

Because of its capacity, a line takes a charging current 
when an alternating E.M.F. is impressed upon it, even 
though it be not connected to a load. The value of this 
current is 

1 = 2 tz/EC 10" 6 amperes, 

where E and C are the voltage and the capacity in micro- 



316 ALTERNATING-CURRENT MACHINES. 

farads respectively between conductors or between one 
conductor and the neutral plane. 

To determine the charging current per conductor of a 
three-phase line, the above formula is used, C and £ being 
the capacity and voltage respectively between conductor 

and neutral plane. The voltage is — - =■ of that between 

conductors. 

The capacity between a conductor and ground may be 
derived by considering the ground as the neutral plane. 
In a circuit not employing a ground return, there is no 
charging current due to the capacity between the con- 
ductors and ground, but in the case of a single conductor 
with ground return there is a charging current. The 
value of this current is given by the preceding formula, in 
which case C is the capacity to ground and E is the voltage 
to ground. 

122. Regulation. — The regulation of a transmission 
line is the ratio of the maximum voltage difference at the 
receiving end, between rated non-inductive load and no- 
load, to the rated-load voltage at the receiving end, con- 
stant voltage being impressed upon the sending end. 

In short aerial transmission lines, the capacity and 
charging current may be considered as negligible. The 
voltage at the receiving end will then be the same as that 
at the sending end on no-load. On full-load the sending 
voltage is equal to the vectorial sum of the delivered 
voltage, that necessary to overcome the resistance drop, 
and that necessary to overcome the inductive drop at 90 
ahead of the delivered voltage. 
. In long transmission lines the capacity cannot be 



POWER TRANSMISSION. 317 

neglected, and the following method due to Steinmetz 
may be employed: Consider the line to be made up of a 
number of sections, say ten, to each of which is appor- 
tioned one-tenth the total capacity and inductance of the 
line. The capacities may be considered as localized 
condensers at the sending end of the section and 
connected across the lines. The inductances may be 
considered as connected in series with the line. The con- 
nections of a few sections are shown in Fig. 224. The 
inductive and resistance drop of voltage in each section is 



ajOT^-r-^TOs^-i r-nms^-r-sim^ 

Sending M 10 I) 9 2 MJ 1 Beceiying 

^OOT^-L^rar^-I L^^5^J_^^3^ 



Fig. 224. 

greater than in any other section more remote from the 
sending end, because, although the inductances and 
resistances are the same in all sections, the current is 
greater as a result of the extra charging current due to 
the capacity of intervening sections. The charging cur- 
rent is also different for each section because the voltage 
which occasions it increases as the sending end is 
approached. 

The voltages and currents in each section can be 
determined with sufficient accuracy by making use of a 
large vector diagram, such as indicated in Fig. 225, where 
OE and 01 represent the delivered voltage and current 
respectively. The power factor of the load being unity, the 
latter are in phase with each other. Let E } E t , E 2 , ... be 
the voltages, and /, I lt I 2 , . . . be the currents delivered 
to the load and the successive sections respectively. Then, 



3i» 



ALTERNATING-CURRENT MACHINES. 



if R and X be the resistance and inductive reactance 
in ohms, and C be the capacity in farads, of each and 
every section, 

E t = (E + RI) XI at 90 lead, 
E 2 = (E t + RI t ) XI t at 90 lead, etc., 
and I x = I coEiC at 90 lead, 

J 2 = /i© ojE 2 C at 90 lead, etc. 

The various phase relations and magnitudes are seen in 
the figure, where the £'s and Fs with various subscripts 




Fig. 225. 



mark the terminals of the vectors from the origin, which 
are not drawn for the sake of clearness. 

The cosine of <j> L , the angle between the current /' and 
the voltage E f at the sending end of the line, is the power 
factor of the line. 

This method is strictly accurate only when the number 
of sections is infinite. Ten sections give sufficient accu- 
racy for practical work. 

123. Conductor Material. — The high permeability of 
iron prohibits its use as a conductor for transmission 



POWER TRANSMISSION. 



319 



lines. There are but two other materials available, copper 
and aluminum. The physical constants of these metals 
are: 



Specific gravity ........ 

Conductivity, in terms of Matthiessen's Standard.. . 

Tensile strength, pounds per square inch 

Elastic limit, pounds per square inch ............. 

Stretch modulus of elasticity, pounds per square inch 
Coefficient of expansion per degree Fahrenheit ..... 



Copper. 


Aluminum. 


8.93 


2.68 


.98 


.61 


60,000 


26,000 


40,000 


14,000 


16,000,000 


9,000,000 


0.0000096 


O.OOOOI28 



For the same conductivity, an aluminum conductor 

must have ~~ or 1.6 times the area of a copper conductor. 
.61 

But as aluminum is three-tenths as heavy as copper, the 

former weighs but 0.48 as much as the latter for equal 

conductivities. Therefore, if aluminum costs less than 

2.08 times as much as copper per unit of weight, it is 

cheaper than the latter. The prices of both metals vary, 

but that of aluminum is usually much less than 2.08 times 

that of copper. If its lower tensile strength and greater 

coefficient of expansion are properly allowed for while the 

line is being strung, the use of aluminum for transmission 

line conductors is just about as satisfactory as that of 

copper. Because of the resultant saving, aluminum is 

being used very extensively. 

124. Insulators. — There is no material from which 
insulators can be made that possesses all the mechanical 
and electrical qualities to be desired. Glass has been used 
to some extent, but it is easily broken, either in transit, or 



320 



ALTERNATING-CURRENT MACHINES. 



by stones and bullets after the insulators are installed. 
The two materials now used for insulators on high-tension 
transmission lines are porcelain and a substance known as 
electrose. 

Porcelain is much tougher than glass and is therefore 
not so liable to be broken. Porcelain insulators are 
heavily glazed to prevent absorption of moisture, since 




Fig. 226. 



even the best porcelain is somewhat porous. A brown or 
gray tint is usually introduced into the glaze so that the 
insulators will not attract the attention of marksmen. A 
Thomas 33 ,000-volt porcelain pin-type insulator is shown 
in Fig. 226. It is 7J inches high and 8| inches in diameter. 
Electrose possesses good insulating qualities, is very 
strong mechanically, and is free from cracks. It has a 
brown, smooth, polished surface and does not absorb 



POWER TRANSMISSION. 



321 



moisture. Metal parts may be molded into it readily if 
so required. A 24,000-volt electrose pin-type insulator is 
shown in section in Fig. 227. It is 7 inches high and 12 
inches in diameter. 

In designing an insulator, the distance along the surface 




227. 



from the conductor to the point of support should be as 
long as possible in order to decrease the leakage current. 
The shortest distance through the air between these two 
points should be great enough to prevent the line voltage 
flashing over, even when the insulator is wet. Further- 
more the distance through the dielectric should be great 



322 ALTERNATING-CURRENT MACHINES. 

enough to prevent puncture. At the same time the size of 
the insulator and the quantity of material used should be 
kept as small as possible. In view of these requirements 
high-voltage insulators have taken the form of a series 
of umbrella-shaped petticoats. Manufacturing difficulties 
prevent the construction of a satisfactory large porcelain 
insulator of the usual type in a single piece. The petti- 
coats therefore are made separately and are glazed or 
cemented together, this being usually done where the line 
is being erected. Electrose insulators may be molded in 
one piece, regardless of shape. 

Metal pins are generally used for pin-type insulators, 
even on wooden poles, because of their strength and the 
fact that they are not burned by the leakage current. It 
was formerly thought that the additional insulation of a 
wooden pin was desirable, but it has been found that even 
the best treated pins will absorb moisture in time, especially 
in salt atmospheres. With steel towers and long spans, 
the large strains on the insulators necessitate the use of 
heavy iron or steel pins. 

The tendency toward higher voltages for power trans- 
mission has led to the application of flexible suspension- 
type insulators. These consist of individual units securely 
coupled together, the number of units to be employed 
depending upon the line voltage. Some 110,000-volt 
insulators of this type are shown attached to a trans- 
mission tower in Fig. 230. 

125. Sag of Conductors. — In stringing the conductors 
there are two things to be taken into consideration. 
First, the greatest possible tension in the conductor, which 
will occur at minimum temperature, must be less than the 



POWER TRANSMISSION. 323 

elastic limit of the conductor. At the minimum tempera- 
ture there may be a coating of ice on the conductor. This 
not only results in increased weight, but also presents a 
greater area to the wind. Second, the clearance between 
conductors and ground at the maximum temperature 
must be great enough to prevent accidental contact or 
malicious interference. These questions are of greater 
importance when using aluminum than when using copper, 
because the tensile strength of aluminum is less than that 
of copper, its coefficient of expansion is greater, and, for a 
given conductivity, it presents a larger surface to the wind. 
The conductors are of necessity strung under varying 
conditions as to temperature, and consequently they 
should be given an appropriate sag so that at the extreme 
temperatures the required conditions will be fulfilled. It 
is common to make use of curves, one for each span 
length, whose ordinates represent the appropriate sag and 
whose abscissae represent temperatures between, say, 
— 40 and no°F. There is considerable difference of 
opinion as to the proper assumptions to be made for sleet 
and wind pressure. In northern countries, conductors 
will frequently be covered with ice from one-half to one 
inch thick all around, and hence this should be allowed 
for. In regard to wind pressure, it should be noted that 
the wind velocities published by the United States Weather 
Bureau are observed and not actual velocities. An 
observed velocity of 100 miles per hour corresponds to an 
actual velocity of about 75 miles per hour. The wind 
pressure in pounds per square foot exerted upon a plane 
surface normal to the direction of the wind may be 
expressed as CV 2 , where V is the actual wind velocity in 
miles per hour and C is a constant whose value may be 



324 



ALTERNATING-CURRENT MACHINES. 



taken as .005. Thus the wind pressure on a plane normal 
surface for an actual velocity of 75 miles per hour is 
approximately 30 pounds per square foot. The wind 
pressure on the conductors is usually taken as half of this 
value, or 15 pounds per square foot of projected conductor 
area. Except in the case of tornadoes, observed wind 
velocities in excess of 100 miles per hour are practically 
unknown. The assumption of a wind pressure of 15 
pounds per square foot of projected area when the con- 
ductor is covered with one-half inch of ice all around is 
conservative. 

The weight of the conductor and ice acts vertically 
downward, while the wind pressure at worst acts trans- 
versely to the direction of the line. The sag of the con- 
ductor is therefore in the direction of the resultant of these 




WEIGHT OF 

CONDUCTOR 

AND ICE 

Fi2. 228. 



5i WIND 
PRESSURE 



two forces, as shown in Fig. 228. The appropriate sag of 
a conductor at any given temperature such that the elastic 



POWER TRANSMISSION. 325 

limit of the metal shall not be exceeded at the minimum 
temperature may be found in the following manner: 
Let S t = span in feet (Fig. 229), 
D = sag in feet (Fig. 228), 

W = weight of conductor and ice in pounds per foot, 
W r = resultant of W and wind pressure in pounds 

per foot, 
T = maximum allowable tension in the conductor, 

usually taken as the elastic limit, 
t = temperature in degrees Fahr. above the mini- 
mum ( — 40°), 
k = temperature coefficient of linear expansion per 

degree F. 
A = cross-section of conductor in square inches, 
E t = stretch modulus of elasticity in pound -inch 

units, 
Ls = length of single span of strung cable at the 

minimum temperature in feet, and 
L u = length of unstressed single span of cable at 
minimum temperature. 



!*- 



-S, J 





Fig. 229. 

Then the following relations are sufficiently exact : 



^-^ + 24 T * 



326 ALTERNATING-CURRENT MACHINES. 

L L > 

D 3 - *£* [L. (i + it) - S,]D = *^ Lu Y" ' 

from which D, the sag in the direction indicated in Fig. 228, 

DW 
may be found. The vertical sag, D' ', is equal to — - • 

w r 

126. Line Structure. — There are two types of line 
structure for transmission lines carrying large amounts of 
power at high voltages, namely, wooden poles and steel 
towers. If wooden poles are used the spans must be 
short, in order that the poles may withstand the forces to 
which they are subjected. With short spans, the expense 
for insulators will be large. On the other hand, while a 
single steel tower costs a great deal more than a wooden 
pole, the use of towers permits of longer spans and 
thereby reduces the number of structures and the cost of 
insulators. The depreciation and repair charges for steel 
towers are very small compared with the same items for 
wooden poles, especially if the towers be galvanized. 
Taking account of interest on the investment and the 
depreciation and repair charges on the line structures, 
including insulators, it will usually be found that towers 
are cheaper than wooden poles for lines using heavy con- 
ductors. Even if the cost of a proposed line with towers 
is a little more than with poles, towers would nevertheless 
be used in most cases because of their greater reliability. 
Towers can be designed to withstand the maximum forces 
which will be exerted upon them, allowing any desired 



POWER TRANSMISSION. 



327 




Fig. 230. 



328 ALTERNATING-CURRENT MACHINES. 

factor of safety. The strength of a tower will remain 
constant, whereas the strength of poles of a given size 
and kind of wood will vary, and the original strength will 
gradually diminish until the poles fail. 

Fig. 230 shows the transmission line tower used by the 
Hydro-Electric Power Commission of Ontario, Canada, in 
transmitting power from Niagara to various points in that 
province at 110,000 volts and 25 cycles. The line con- 
sists of two three-phase circuits, the cables of each circuit 
being nine feet apart. The tower is 60 feet high, with a 
base 16 feet square, the span being 550 feet. Lightning 
protection is secured by the use of overhead ground wires 
fastened to the tower, as shown. 

The forces acting on a structure in a straight portion of 
the line, where the spans are of equal length, are: (a) the 
weight of the conductors and ice; (b) the wind pressure 
on the conductors when covered with ice; (c) the wind 
pressure on the structure; (d) the weight of the structure. 

The method of calculating the first three forces has been 
considered. The first and last act vertically downward, 
while the others are considered as acting transversely to 
the direction of the line. The value of the wind pressure 
is a maximum when the direction of the wind is trans- 
verse to that of the line, and zero when the wind is in the 
same direction as the line. Theoretically, there is no 
force on the structure due to the tension in the conductors, 
since the tensions in two adjacent spans counterbalance 
each other. In practice, however, such a force exists, on 
account of differences of wind pressure, of elevation of the 
towers, and of length of spans. 

Should one or more of the conductors break, the full 
tension thereof will come on the structures on each side of 



POWER TRANSMISSION. 329 

the break. Such an accident is unusual, and the addi- 
tional expense of building each structure to withstand it 
would not be justified. Consequently the conductors are 
secured to the insulators by means of loose ties. In the 
event of the breaking of a conductor the tension will then 
be taken up by several structures. At intervals of a few 
miles there are guyed towers capable of taking the full 
tension, and the conductors are firmly clamped to the 
insulators on such structures. 

Knowing the magnitude of the transverse forces, the 
strength of the pole required to resist them can readily be 
calculated by means of the well-known formula for a beam 
fixed at one end. The resisting moment 

M - c> 

where 5 is the stress in the section, / is the moment of 
inertia of the section, and C is the distance from center to 
the fiber under maximum stress. From this formula, the 
most economical wooden pole is one whose vertical sec- 
tion is a parabola. The strength of the pole necessary 
to resist the other two forces can be determined by means 
of a formula for the compressive strength of a long 
column. However, if a structure is strong enough to 
resist the transverse forces, it will undoubtedly be strong 
enough to resist the vertical forces. 

127. Spans and Layout. — For a given sized conductor, 
the necessary heights of the line structure, and the loads 
which the latter must sustain, increase as the span is 
lengthened. Consequently the cost of a single structure 
increases as the span is increased. At the same time, 
however, the number of structures and of insulators 



33o 



ALTERNATING-CURRENT MACHINES. 



required is diminished. If the total cost of the structures 
complete with insulators be calculated for different spans, 
the most economical length of span can be determined. 
The foregoing applies to either wooden poles or steel 
towers, but of course poles can be used only for com- 
paratively short spans because of strength limitations. 

On account of irregularities in the ground and in order 
to clear obstacles, spans longer or shorter than the stand- 
ard are frequently necessary. In such cases the forces 
acting on the structures are other than normal. These 
may sometimes be taken care of by using the regular 
structures with guys, but otherwise special structures are 
necessary. 

In straight portions of the line ordinarily there are no 
forces acting on the structures due to the tension in the 

conductors, since the 

tension on one side 

balances that on the 

other. But when the 

line changes in direction 

this is not so. Figs. 231 and 232 show two 

methods of making a change in direction. 

In the first, the conductors are dead-ended 

on the two structures, and the tension is taken 

up by guys. In Fig. 232, for simplicity, 

but one conductor is shown. The effect of 

Fig. 231. 

the tension in the conductors on the struc- 
tures in changing the direction of the line by this 
method is shown in Fig. 233. For each conductor 
there is a force acting transversely on the structure 

equal to 2 sin — times the tension in the conductor. The 




POWER TRANSMISSION. 



331 



forces acting on structures at curves should be the same as 
on straight portions. Therefore the spans on curves are 
shortened by such an amount that the sum of the wind 




Fig. 232. 



pressure on the conductors and the transverse component 
of the tension shall be equal to the wind pressure on the 
conductors on straight portions of the line. With shorter 




Fig. 233. 

spans the tension in the conductors is less, and therefore 
the structures A and F, Fig. 232, must be guyed in order 
to equalize the tensions. 

1 28. Example of Design of Transmission Line. — Let it be 

required to transmit 5000 kilowatts a distance of 100 miles 
over a three-phase circuit, using aluminum conductors, 
with a voltage of 66,000 at the generating station, the 
frequency being 25. 

Economic Drop. The formulae for the economic drop 
and cross-section of conductor for a three-phase line are 
given in § 118 as equations (7) and (8) respectively. Since 
the conductors are to be of aluminum, 
K t = 85,000 ohms, 
K 2 = 0.0048 pound, 



332 ALTERNATING-CURRENT MACHINES. 

as calculated from the constants given in § 123. Assuming 

c t = $15.00, 
c 2 = $ 0.25, 

P2 = O.O5, 

then K = 4 X .05 X .25 X 85,000 X .0048 X 10,000 X 1000 
= 204,000,000. 

Hence x = - I^L^poo^ooo 

4 (66,ooo) 2 X 15 
fei^4^co 2 oooJM 1 i2j6^^ 

4 (66,ooo) 2 x 15 =.0461, 

or the economic drop is 4.61 % of the impressed voltage. 

If 5000 k.w. are to be delivered, then — ^2P__ = 

1 — .0461 ^ 
must be supplied to the line. The size of the conductor 

required is 

5 = if"iooo2<_5241o, 200 1 

c 0^,000 

2 U;35 6 ,ooo,ooo J} .0461J 
= 221,900 circular mils. 

An aluminum conductor of this size weighs 221,900 X.0048 
or 1065 pounds per mile. 

Proof that 4.61 % is the economic drop for the a ive n 
conditions : & 

(a) Annual cost of lost power = 242 X $is = $3630. 

(b) Interest on cost of line conductors = 1065 x'300 

X $0.25 X .05 = $3994. 

(c) Cost of lost power per delivered kilowatt-year 

. - 3 630 (t]> ^ ,r 

— 5000 = $0,726. 

(d) Interest on cost of line conductors per delivered 
kilowatt-year = UU = $0,799. 



POWER TRANSMISSION. 333 

Hence the sum of loss and interest per delivered kilowatt- 
year is $1,525. 

Now suppose | as much conductor material were used. 

Line loss = f X 242 = 277 k.w. 

Drop = f X 4.61 = 5-27 %• 

Delivered power = 5242 — 277 = 4965 k.w. 

Then the new values of a, b, c, and d are: 

a = 277 X $15 = $4155. 
b = I X $3994 = $3498, 
c = UU = $0,838, 
d = Ifff = $0,704, 

and hence the sum of loss and interest per delivered 
kilowatt-year is $1,542. Therefore using -J as much con- 
ductor material increases the cost of delivering power. 
The cost of lost power and interest on the cost of the line 
conductors have been similarly calculated for f , -f, and f 
as much conductor material. The results, which are 
plotted in Fig. 234, prove that 4.61 % is the economic 
drop, since the cost per delivered kilowatt-year is a mini- 
mum at that value. 

It has been assumed that the conductor is solid wire; 
but for a conductor of this size, cable is always used. 
The resistance of a cable is a few per cent higher than that 
of a solid conductor having the same cross-sectional area. 
A cable of 228,000 circular mils has approximately the 
same resistance as a solid wire of 221,900 circular mils. 
A cable of this size weighs 1 100 pounds per mile, has a resist- 
ance of 0.396 ohm per mile, and is 0.55 inch in diameter. 

Inductance. — Assuming the conductors to be spaced 



334 



ALTERNATING-CURRENT MACHINES. 



72 inches apart, center to center, the inductance per mile 
of two conductors 0.55 inch in diameter, § 120, is 

2 80.5 + 740 log -^- io~ 6 = 0.00374 henry, 
L - 2 75J 

and the inductance of the whole length of the line, for two 
conductors, is 0.374 henry. 

Capacity. From § 121, the capacity between two 

conductors 0.55 inch in diameter is ° ,OI 94 = 0.00803 

2.418 

microfarad per mile, and the capacity between two con- 
ductors for the whole length of line is 0.803 microfarad. 





























-YEAR 

83 


L 
























\ 
























IVERED KW.- 

s 


\ 
















































UJ 

O 

K .1.54 

<r> 












































































O 

° L52 




























1.50 



























4 5 

DROP. IN PER CENT OF IMPRESSED E. M. F. 
Fig- 234. 



Regulation. A convenient way of calculating the 
regulation of a three-phase circuit is based upon the fact 
that a three-phase circuit is equivalent to two single- 



POWER TRANSMISSION, 335 

phase circuits employing conductors of the same size. In 
other words, the regulation of a three-phase circuit is the 
same as that of a single-phase circuit carrying half as 
much power with the same percentage loss, at the same 
voltage and distance between conductors. The induc- 
tance and capacity of a single-phase line with the same 
sized conductors as the three-phase line under considera- 
tion have just been calculated. Dividing the line into ten 
equal sections, the constants of each are 

L = .0374 henry. 

C = .0803 microfarad. 

R = .396 X 20 = 7.92 ohms. 

Using the notation of § 122, and assuming the voltage at 
the receiving end of the line 

E = 66,000 (1 — .0461) =-■ 62,957 v °lts. 

If 2500 kilowatts are delivered, the load current is 

T 2,^00,000 

/ = -~ — = SQ-7I amperes. 

The resistance and reactance drops of section 1 are 
respectively 

IR = 39.71 X 7.92 = 314 volts, 
2 nfLI = 50 Ti .0374 X 39.71 = 233 volts. 

Hence 

E t = V (62,957 + 3^7 + (233)' = 63,271.4 volts. 

The current in section 1 is 

I x = v / (39 > 7i) 2 + (507163,271.4 X .0000000803 ) 2 
= 39.72 amperes. 



336 ALTERNATING-CURRENT MACHINES. 

Similarly, the E.M.F. across section 2 is 

E 2 =V(6 3 , 271.4 + 39.72 X 7-92) 2 + (50 n .0374 X 39-72) 2 
= 63,586.0 volts, 

and the current therein is 



I 2 = v (39»72) 2 + (5071 63, 586 X. 0000000803 ) 2 = 39.73 amps. 

Proceeding in like manner, the values of the E.M.F.'s and 
currents in each section may be determined, and the regu- 
lation then calculated. 

Natural Frequency. The inductance of the line is 
0.374 henry and the capacity is 0.000000803 farad. From 
§115, the natural frequency is 

= 456 cycles. 



4 V 0.374 X .000000803 

There is therefore no probability of trouble from har- 
monics at the chosen frequency of 25. 

Sag of Conductor. The conductors are to be strung 
with such a sag that at the minimum temperature with 
one-half inch of ice all around the cable, and a wind pres- 
sure of 15 pounds per square foot of projected area, the 
tension in the cable shall not exceed the elastic limit of the 
material (14,000 pounds per square inch). 

Weight per foot of cable is Hlo = 0.208 pound. 

Area of conductor is 0.179 square inch. 

Outside diameter of cable is 0.55 inch. 

Since ice weighs 57 pounds per cubic foot, the weight of 
an ice coating one-half inch thick is 



w- (f 



12 n — — X 57 = 0.652 pound per foot of cable. 



POWER TRANSMISSION. 337 

The weight of cable and ice is therefore 0.86 pound per 
foot of cable. 

The wind pressure on the ice-covered cable is 

—^ X 15 = 1.94 pounds per foot of cable. 

144 

The resultant of weight and wind pressure is 



\/(o.86) 2 + (1.94) 2 = 2.122 pounds per foot of cable. 

Assuming a span of 400 feet, then in the notation of 
§125, 

S t = 400. 
W = 0.86. 

W r = 2.122. 

A = 0.179. 

T = 14,000 X 0.179 = 2 5 10 - 

t = o, 75, and 150. 

k = 0.0000128. 
E ! = 9,000,000. 

Hence 

(4Q0) 3 X(2.I22) 2 
24 (25IO) 2 



L s = 400 + ■* — ^ ,_ x 2 = 401.91, 



r 401.91 

L u = > - = 401.28. 

2 5 IQ 
1 + - 



9,000,000 X 0.179 
If / = o, then 

g3 _3_X4go g _ = 3 X(4oo) 3 X40i. 2 8X2.i22 ; 

8 64 X 9,000,000 X 0.179 

or D B — 192 D = 1585.7 



338 ALTERNATING-CURRENT MACHINES. 

Solving by estimation and trial, the sag, D, is found to be 
16.9 feet. 

The vertical sag D' = — ' — X 16.0 = 6.84 feet. 
2.122 

If t = 75, there results 

D 3 - 3 X o 4 °° [401.28 (1 + 0.00096) - 400] D = 1585.7, 
8 

from which D = 18.35 ^ eet - 

The vertical sag = — - — X 18.35 = 7.43 feet. 
2.122 

If / = 150, then 

£> 3 - 3 X Q 4 °° [401.28 (1 + 0.00192)- 400] D= 1585.7; 
8 

.'. D = 19.69 feet. 

Vertical sag = — — X 19.69 = 7.98 feet. 
2.122 

If the minimum temperature is taken as — 40 F., 75 
above the minimum is 35 F., and 150 above the mini- 
mum is no° F. In Fig. 235 the vertical sags for spans of 
400 feet, 500 feet, 600 feet, and 700 feet have been plotted 
in terms of temperatures between — 40 and no°F. In 
stringing the cables, the proper sag to be allowed should 
be obtained from these curves, its value depending upon 
the temperature at that time. 

Length of Standard Span. From the curves of 
Fig. 235, the lower curve of Fig. 236 has been drawn, 
showing the vertical sag at no° F. for different span 
lengths. If the minimum clearance of the cables from 
the ground is to be 20 feet, the point of support of the 



POWER TRANSMISSION 



339 



cables must be at a distance from the ground equal to 20 
feet plus the maximum sag. The distance of the point of 
support of the lowest cable from the ground is called, for 



22 










































__70C 


iFTJ 


*PAN 
















20 






























































18 












































































600 


FT.? 


(PAN 














16 
































Id 

Z 14 



<c 
to 






























































-1 
< 


few 

UJ 

> 












































500 


FT.S 


=AN- 














10 






























































8 










































4 


)OFT 


..SPW 


j 














6 








































1 



















-40 c 



20° 40 u 

TEMPERATURE{FAHR.) 

Fig. 235. 



80 : 



.100 ^ 



convenience, the height of the tower. The upper curve of 
Fig. 236 has been drawn with ordinates representing 20 
feet more than those of the lower one, and therefore shows 



34Q 



ALTERNATING-CURRENT MACHINES. 



the heights of towers for different span lengths. Assume 
that 66,ooo-volt insulators cost $5.00 each erected, and that 



10 
36 

32 
28 
24 


























































-fO* 


^2 


G^5- 




















































































18 
12 


































s££ 


£^ 


cTj>- 




































8 
4 



























400 



500 600 

SPAN LENGTH IN FEET 

Fig. 236. 



700 



the costs of towers of various heights, erected, are as 
follows : 



Tower Height in Feet. 


Cost of Tower in Dollars. 


3° 

32-5 

35 

37-5 
40 


95 
100 
no 
125 
145 



From the upper curve of Fig. 236, it is seen that the 
greatest span length for which a 30-foot tower can be used 
under the given conditions is 450 feet. With this span 



POWER TRANSMISSION. 341 

length there will be required 11.73 towers per mile, and 

the cost of towers and insulators per mile of line will be 

11.73 X $95 = $1114.35 for towers. 

3 X n.73 X $5 = 175.95 for insulators. 

$1290.30 = total cost per mile. 

For 32.5-foot towers the span is 515 feet, and 10.25 
towers are required per mile. Then 

10.25 X$ioo = $1025.00 for towers. 
3 X 10.25 X $5 = 153-75 f° r insulators. 

$1178.75 = total cost per mile. 

For 35-foot towers the span is 570 feet, and 9.26 towers 
are required per mile. Then 

9.26 X $110 = $1018.60 for towers. 
3 X 9.26 X $5 = 138.90 for insulators. 

$1157.50 = total cost per mile. 

For 37.5-foot towers the span is 615 feet, and 8.6 towers 
are required per mile. Then 

8.6 X $125 = $1075.00 for towers. 
3 X 8.6 X $5 = 129.00 for insulators. 

$1204.00 = total cost per mile. 

For 40-foot towers the span is 665 feet, and 7.94 towers 
are required per mile. Then 

7.94 X $145 = $1151.30 for towers. 
3 X 7.94 X $5 = 1 19. 10 for insulators. 

$1270.40 = total cost per mile. 

It is evident from the foregoing calculations that the 
economic span is 570 feet, employing 3 5 -foot towers. 
Forces Acting on the Towers. For spans of 570 feet, 



342 ALTERNATING-CURRENT MACHINES. 

the force acting on each tower due to the weight of line 
conductors when covered with ice will be 

3 X 570 X 0.86 = 1470.6 pounds. 

The pressure due to the wind, being 15 pounds per 
square foot of projected cable area, when the cable is 
covered with one-half inch of ice all around, is 
3 X 570 X 1.94 = 33 J 7-4 pounds. 

The weights of towers vary considerably, depending 
upon their design. One ton may be taken as the average 
weight of a 35-foot tower. 

A tower of the size under consideration will have the 
equivalent of about 25 square feet of normal surface 
exposed to the wind. Hence the wind pressure on the 
tower is 25 X 30 = 750 pounds. This acts at the center 
of gravity of the exposed surface, but for the purpose of 
calculation it is assumed that half this force, 375 pounds, 
acts at the top of the tower. 

Therefore the tower must be strong enough to resist a 
force of 1470 + 2000 = 3470 pounds acting vertically 
downward, and a force of 3317 + 375 = 3692 pounds 
acting horizontally at the top of the tower. 

Length of Span on Curves. Where the line is carried 
around a curve as shown in Fig. 232, the transverse force 
acting on the tower due to the tension in the cables 
should be allowed for by shortening the span length. 
If the angle a be 2 , the transverse force due to the tension 
in the cables (Fig. 233) is 

3 X 2 X 2510 X sin i° = 263.5 pounds.. 

The transverse force due to wind pressure on the con- 
ductors is 3317.4 pounds in the standard span. Sub- 
tracting 263.5 therefrom leaves 3054 pounds as the 



POWER TRANSMISSION, 343 

desired wind pressure on the conductors per span on the 
curve. Hence the length of such spans should be 

^^-X 57° = 5 2 5 f eet. 
33*7 

If the angle a be 4 , then the transverse force due to 
the tension in the cables is 525.6 pounds. The span length 
for this value of a is 

3317 ' 4 - 525 ^ X 570 = 480 feet. 
33 J 7 

Similarly, when 

a = 6°, the span is 432 feet, 

a = 8°, the span is 390 feet, 

a = io°, the span is 345 feet. 

It is not advisable to have the angle a greater than io°. 
If too many towers will then be required to make the 
necessary turn, it is better to make it as shown in Fig. 231, 
by dead-ending the line on two towers and having a short 
slack span between them, rather than by means of a curve. 

PROBLEM. 

Thirty thousand kilowatts are to be transmitted over a section of a 
transmission line 53 miles long, using a three-phase circuit of alumi- 
num conductors, with 110,000 volts at the generating station. The 
various constants are: 

Frequency = 25. 

Cost of power per kilowatt-year at generating station = $12.00. 

Cost of aluminum per pound = $0.25. 

Interest rate thereon = 4 %. 

Distance between cables = 9 feet. 

(Fig. 230 shows the type of towers used.) 

Determine the economic drop, cross-section of conductor, natural 
frequency of the line, and the charging current per conductor. Pre- 
pare curves showing the vertical sag at different temperatures for 
various span lengths. 



INDEX. 



[The figures refer to page numbers.] 



Addition of vectors, 17. 
Admittance of circuit, 72. 

representation of, 74, 
Admittances, polygon of, 87. 
Ageing of iron, 188. 
Air-blast transformers, 194. 
Air gap of induction motors, 227. 
Alexanderson alternator^ 145. 
All-day efficiency, 167= 
Alternating current, definition of, 1 

power transmission, 300. 
Alternations, definition of, 1. 
Alternator, 94. 

compensated, 128, 

efficiency of, 133. 

flux in, 1 170 

General Electric Co/s, 125, 128, 
141. 

inductor type, 136. 

losses in, 134. 

rating of, 135. 

regulation, 116. 

revolving-field type, 139. 

saturation curves, 114. 

self-exciting, 145, 

Stanley, 137. 

voltage drop in, 117. 

voltage of, 96. 

Westinghouse, 126. 
Alternators in parallel, 262. 
Aluminum line wire, 319. 
Angle of hysteretic advance, 160. 
Angle of lag or lead, 12, 71. 



Apparent resistance, 38, 72. 
Armature copper ioss, 120, 134. 

E.M.F Q generated in, 96. 
Armature impedance voltage, 121. 

inductance, 119. 

reaction of converters, 29 i„ 

resistance drop, I2i„ 

windings, 99. 
Autotransformer, 151. 

connections of, 186. 
Average value of current and pres- 
sure, 8. 

Balanced polyphase systems, 105. 
Belt leakage reactance, 224. 

Calculation of alternator regulation, 
119. 
induction motor leakage react- 
ance, 216 
resultant admittance, 89, 91. 

impedance, 86, 91. 
transformer leakage inductance, 
168. 
Capacity, distributed, 303. 
formulas, 53. 
of condensers, 50. 
of transmission lines, 313, 334. 
reactance, 64, 72, 
unit of, 51. 
Centrifugal clutch pulley, 210. 
Charging current of transmission 
line, 315. 



345 



346 



INDEX. 



Choke coils, 44. 

Circle diagram of induction motor, 
231, 233. 
of transformer, 179. 
Circuits, natural period of, 80, 83. 

time constant of, 34. 

with R, L, and C, 70. 
Coefficient, leakage, 233. 

of self-induction, 27. 
Coil-end leakage reactance, 222. 
Compensated alternators, 128. 

series motors, 270. 
Compensators, connections of, 186. 

synchronous, 257. 
Complex numbers, representation of 

Z and Y by, 74. 
Composite winding, 125. 
Concentrated armature windings, 99. 
Condenser, capacity of, 50. 

compensator, 241. 

construction of, 51. 

hydraulic analogy, 60. 

resistance, 52. 
Condensers, 48. 

in parallel and in series, 55. 
Condensive circuit, phase relations 

in, 61. 
Conductance of circuit, 73. 
Conductive compensation, 271. 
Connections of transformers, 181. 
Constant -current transformers, 195. 

potential, regulation for, 124. 
Converter, 284. 

armature heating, 290. 
reaction, 291. 

capacity, 291. 

coils, current in, 290. 

current relations in, 288. 

E.M.F. relations in, 286. 

hunting of, 293. 

inverted, 285. 

mercury vapor, 296. 



Converter, regulation of, 293. 
split-pole, 296. 
starting of, 291. 
Cooling of transformers, 192. 
Copper line wire, 319. 
loss in transformers, 165. 
of armature, 120, 134. 
Core flux of transformers, 154. 
loss in transformers, 156. 
-type of transformer, 149, 191. 
Counter E.M.F. of self-induction, 27. 
Critical frequency, 81. 
Current and voltage relations in 
condensive circuit, 63* 
in polyphase systems, 101, 104, 
106. 
average value of, 8. 
components of, 16, 159, 227. 
effective value of, 7. 
flow, expression for, 71. 
instantaneous, in alternating-cur- 
rent circuits, 41, 65, 76. 
values of, 3. 
lag or lead of, 12. 
magnetic energy of started, 36. 
produced by harmonic E.M.F. , 

37, 64. 
relations in converters, 288. 
Currents, single-phase and poly- 
phase, 13. 
Curve, efficiency, of alternator, 135. 
transformer, 167. 
non-sine, form factor of, 10. 
saturation, 114. 
sine, 4. 

form factor of, 9. 
Curves in transmission line, 330. 
Cycle, definition of, 1. 

Damped oscillations, 82. 

effective current value of, 83 
Damping factor, 82. 



INDEX. 



347 



Decaying currents, 34, 57. 

oscillatory current, 82. 
Decrement of oscillations, 82. 
Definition of terms, 71. 
Delta connection, 100, 184. 
Design of transmission line, 331. 
Dielectric constants, 52. 

energy stored in, 60. 

for condensers, 51. 

hysteresis, 52. 

polarization, E.M.F. of, 58. 

strength of materials, 50. 
Direct-current power transmission, 

300. 
Distance between line conductors, 

3*3° 
Distortion of E.M.F. wave, causes 

m of, 5- 
Distributed capacity, 303. 

windings, 97, 101, 205, 272. 
Distribution constant, 98. 
Drop of voltage in transmission 
lines, 306, 331. 

Economic drop in line, 306, 331. 
Eddy current loss in induction mo- 
tors, 228. 
transformers, 157. 
Effective values of current and 

pressure, 7. 
Efficiency, all-day, 167. 
curve of alternators, 135. 
of alternators, 133. 
induction motors, 214. 
transformers, 166. 
E.M.F., average value of, 8. 
counter, of self-induction, 27. 
effective value of, 7. 
generated in armature, 96. 
instantaneous value of, 5, 24, 
methods of calculating alternator 
regulation, 119. 



E.M.F., of dielectric polarization, 58. 
synchronous motor, 256. 
relations in converters, 286. 
wave, shape of, 6. 
E.M.F.'s in series, 20. 

of plain series motor, 265. 
Electrose insulators, 321. 
Electrostatic capacity, see Capacity. 
Energy of a started current, 36. 

stored in dielectric, 60. 
Equivalent R, X, and Z of trans- 
former, 163. 
rotor resistance, 235. 
sine wave, definition of, 18. 
Exciting current of induction motor, 
227, 233. 
transformer, 151, 159. 
Expression for current flow in any 
circuit, 71. 

Farad, definition of, 51. 
Field, rotating, 202. 
Flux density in induction motors, 
229. 
transformers, 155. 
fringing constant, 222. 
Forced compensation, 271. 
Form factor, definition of, g. 
of non-sine curves, 10, 
sine curve, 9. 
Formulae for calculating capacities, 

53- 

for calculating inductances, 31. 
Four-phase systems, 106. 
Fractional-pitch motor windings 

218. 
Frequencies, standard, 2, 302. 
Frequency and speed, 2. 

changers, 244. 

for power transmission, 302. 

natural, of transmission lines, 303. 

resonant, 80. 



348 



INDEX. 



Full-load saturation curve, 114. 
-pitch motor windings, 218. 

Gauss, definition of, 28. 
General Electric Co.'s alternator, 
125, 128, 141, 145. 

induction motor, 204. 

motor starter, 208. 

regulator, 295. 

synchronous motor, 259. 

transformer, 190, 194. 
Growth of current in inductive cir- 
cuit, 33. 

Harmonic shadowgraph, 3. 
Harmonics of fundamental E.M.F., 

23- 
Heating of converter coils, 290. 
Henry, definition of, 27, 
Hunting of converters, 293. 

synchronous motors, 256. 
Hydraulic analogy of condenser, 60. 
Hysteresis, dielectric, 52. 

loop, 162. 

loss in induction motors, 228. 
Hysteresis loss in transformers, 158. 
Hysteretic advance, angle of, 160. 

constant, 158. 

Ideal transformer, 151. 

vector diagram of, 154. 
Impedance, definition of, 38. 

of circuit, 72. 

representation of, 74. 

synchronous, 117. 

voltage, armature, 121. 
Impedances in series and in parallel, 
90. 

polygon of, 83. 
Inductance, armature, 119. 

formulae for, 31. 

of transmission lines, 310, 33^- 

practical values of, 29. 



Inductance, self, described, 260 

unit of self, 27. 
Induction motor, 202. 
air gap of, 227. 
calculation of exciting current, 
227. 
of leakage reactance, 216. 
circle diagram of, 231, 233. 
efficiency of, 214, 241. 
exciting current of, 233. 
flux density in, 229. 
General Electric Co.'s, 204. 
leakage coefficient of, 233. 
losses in, 240. 

magnetizing current of, 230. 
performance curves, 236. 
power factor of, 232. 
resistance of windings, 235. 
rotors of, 205. 
single-phase, 242. 
slip of, 210, 241. 
speed and efficiency, 213. 

regulation, 245. 
starting of, 207, 244. 
test with load, 238. 
torque and slip, 212, 226. 
torque of, 214. 
transformer method of treat- 

ment, 215. 
Westinghouse, 204. 
windings, 205, 218. 
wattmeter, 246. 
Inductive compensation, 272. 

reactance, 38, 72. 
Inductor alternators, 136. 
Instantaneous current in alternat- 
ing-current circuits, 41, 65, 76. 
values of current and voltage 4, 
24. 
Insulators, 319. 
Interpretation of symbol /, 75, 
Inverted converter, 285. 



INDEX. 



349 



Lag or lead of current, 12, 71. 
Leakage coefficient, 233. 

reactance of induction motors, 
216. 
transformers, 168. 
Lighting transformers, 188. 
Lightning arrester choke coils, 46. 
Line capacity, 313, 334. 

constants, 319. 
Line inductance, 310, 333. 
natural frequency of, 303. 
resistance, 309. 
structure, 326. 
wire, cross-section of, 307. 
material, 318. 
relative weights of, 305. 
sag of, 322, 336. 
wires, distance between, 313. 
wind pressure on, 323. 
Linkages defined, 27. 
Load losses in alternators, 134. 
saturation curve, 114. 
test on induction motors, 238. 
Logarithmic change of current, 34. 

decrement of oscillations, 82. 
Losses in induction motors, 240. 
synchronous machines, 134. 
transformers, 156. 

Maclaurin's series, 76. 
Magnetic energy of started current, 
36. 
flux in alternators, 117,, 
leakage in induction motor, 216. 
transformer, 168. 
Magnetizing current of induction 
motor, 230. 
transformer, 159. 
wave of transformer, 162. 
M.M.F. method of calculating alter- 
nator regulation, 123. 
Magnitude of self-induction, 30. 



Material of line conductors, 318. 
Maxwell, definition of, 28. 
Measurement of power, 107. 
Mercury vapor converter, 296. 
Mesh or delta connection, 100, 

184. 
Microfarad, definition of, 51. 
Monocyclic system, 244. 
Motor, induction, see Induction mo- 
tor, 
repulsion, 277. 
series-repulsion, 280. 

single-phase, see Series motor, 
starters, General Electric Co.'s, 
207. 
Westinghouse, 209. 
synchronous, see Synchronous 
motor. 

Natural draft transformers, 192. 
frequency of transmission lines, 

3°3- 
period of circuit, 80, 83. 
No-load saturation curve, 114. 
Non-sine curves, form factor of, 10. 
phase difference of, 18. 

Obstructance, definition of, 49. 
Oil-cooled transformers, 194, 
Operation of induction motors, 210. 
Operative range of synchronous 

motors, 253. 
Oscillations, damped, 82. 

Parallelogram of E.M.F.'s, 21. 
Parallel operation of alternators, 

262. 
Percentage of saturation, 114. 
Performance curves of induction 

motor, 236. 
Phase, 12. 

-belt of conductors, 223. 



350 



INDEX. 



Phase, difference of non-sine curves, 
18. 
sine curves, 12. 
or distribution constant, 98. 
relations in condensive circuit, 
61. 
Phases, number of, for transmission, 

3°4- 
Phase splitters, 241. 

-wound rotors, 206. 
Pin-type insulators, 320. 
Pitch factor of induction motors, 

218. 
Plain series motor, 264. 

characteristics of, 268. 
Polygon of admittances, 87. 
E.M.F.'s, 22. 
impedances, 83. 
Polyphase alternators, 94. 
currents, 14. 

power, measurement of, 109. 
transformers, 198. 
Porcelain insulators, 320. 
Power component of current, 16, 
159, 227. 
factor, definition of, 17. 
of induction motor, 232. 
of three-phase balanced circuits, 

112. 
of transmission lines, 318. 
in alternating-current circuits, 14. 
measurement of, 107. 
transmission, frequency for, 302. 
number of phases for, 304. 
systems of, 300. 
voltage for, 305. 
Pressure, average value of, 8. 
Pressure curves, actual, 6. 
distortion of, 5. 
effective value of, 7. 
for power transmission, 305. 
instantaneous value of, 4. 



Preventive leads, 276. 

Primary of induction motor, 216. 

transformer, 149. 
Problems, 25, 47, 68, 92, 146, 200, 
282, 299, 343. 

Quarter-phase currents, 13. 
systems, 101. 

Radius vector, 5. 
Rating of alternators, 135. 
Ratio of transformation, 149. 
Reactance of any circuit, 72. 
condensive circuit, 64, 
inductive circuit, 38. 
Reactors, 44. 

Rectifier, mercury vapor, 296. 
Regulation for constant potential, 
124. 
of alternators, 116. 

methods of calculating, 119. 
of converters, 293. 
of induction motors, 245. 
of transformers, 173, 181. 
of transmission line, 316, 334. 
Regulator, General Electric Co.'s, 

295- 
Stillwell, 294. 
Tirrill, 130. 
Reluctance of transformer core, 

160. 
Representation of Z and Y by com- 
plex numbers, 74. 
Repulsion motor, 277. 

starting of single-phase induc- 
tion motor, 279. 
series motor, 280. 
Resistance, apparent, 38, 72. 
drop, armature, 121. 
leads for series motors, 276. 
of line wire, 309. 
Resonance, 80. 



INDEX. 



351 



Resultant admittance, 89. 

E.M.F. of harmonic components, 

23. 

impedance, 86. 
Revolving-field type alternators, 139. 
Rotary converter, see Converter. 
Rotating magnetic field, 202. 
Rotor of induction motor, 203. 

phase-wound, 206. 

squirrel-cage, 205. 

Sag of transmission lines, 322, 336. 
Saturation, 113. 

curves of alternator, 114. 

factor, 115. 
Scott transformer, 183. 
Secondary of induction motor, 216. 

transformer, 149. 
Self-exciting alternator, 145. 

-inductance, counter E.M.F. of, 

27, 37- 
described, 26. 
formulae for, 31. 
unit of, 27. 
Series motor, compensated, 270. 
characteristics of, 273. 
connections of, 272. 
performance curves, 275. 
plain, 264. 

characteristics of, 268. 
E.M.F.'s of, 265. 
resistance leads of, 276. 
sparking in, 274. 
Westinghouse compensated, 271. 
-repulsion motor, 280. 
Shading coil, 248. 
Shadowgraph, harmonic, 3. 
Shell-type of transformer, 149, 192. 
Sine curve, 4. 

form factor of, 9. 
wave, equivalent, definition of, 
18. 



Single-phase alternators, 94. 

commutator motors, 262. 

current, 13. 

induction motor, 242, 279. 
Skin effect of wire, 45, 310. 
Slip of induction motors, 210, 241. 
Slot contraction factor, 225. 

leakage reactance, 217. 
Solenoids, self-inductance of, 32. 
Span lengths on curves, 342. 
Spans on transmission lines, 329, 

338. 
Sparking in series motors, 274. 
Split-pole converter, 296. 
Squirrel-cage motors, starting of, 
207. 

rotors, 205. 
Standard frequencies, 2. 
Stanley alternator, 137. 

transformer, 192. 
Star or F-connection, 100, 184. 
Started current, magnetic energy of, 

Starting converters, 291. 

induction motors, 207, 244, 279. 

synchronous motors, 258. 
Stator of induction motor, 203. 

windings, 205. 
Step-up and step-down transfor- 
mation, 150. 
Stillwell regulator, 294. 
Strength of dielectrics, 50. 
Structures, transmission line, 326. 
Susceptance of circuit, 73. 
Suspension-type insulators, 322. 
Synchronizer, 258. 
Synchronous compensators, 257. 

converter, see Converter. 

impedance, 117. 

machines, losses in, 134. 

motor, 249. 

behavior of, 252. 



352 



INDEX. 



Synchronous motor: 

efficiency of, 255. 

E.M.F., 256. 

General Electric Co.'s, 259. 

hunting of, 256. 

operative range of, 253. 

stability of, 255. 

starting, 258. 

V-curves of, 258. 
Synchroscope, 260. 

Table of converter capacities, 291. 

dielectric constants, 52. 
strengths, 50. 

line constants, 319. 
Temperature effect on core loss, 159. 
Three-phase power measurement, 
no. 

systems, 104. 

transformations, 184. 
Thury system of power transmis- 
sion, 300. 
Time constant of circuit, 34, 59. 
Tirrill regulator, 130. 
Tooth-tip leakage reactance, 220. 
Torque of induction motors, 214. 
Transformation, ratio of, 149. 
Transformer, air-blast, 194. 

calculation of leakage reactance 
of, 168. 

circle diagram of, 179. 

connections, 181. 

constant-current, 195. 

cooling of, 192. 

copper losses, 165. 

core losses, 156. 
reluctance of, 160. 

definitions, 149. 

eddy current loss in, 157. 

efficiency, 166. 

equivalent R and X of, 163. 

exciting current of, 151, 159. 



Transformer, flux in, 154. 
for lighting, 188. 
General Electric Co.'s, 190, 194, 

197. 
graphic representation of, 151. 
hysteresis loss in, 158. 
ideal, 151. 

vector diagram of, 154. 
losses, 156. 
magnetizing current of, 159. 

wave, 162. 
method of induction motor treat- 
ment, 215. 
oil-cooled, 194. 
polyphase, 198. 
regulation of, 173, 181. 
Scott, 183. 
Stanley, 192. 

vector diagram of, 175, 178. 
Wagner, 188. 
water-cooled, 194. 
Westinghouse, 191. 
with divided coils, 172. 
Transmission line, see also Line, 
charging current of, 315. 
design of, 331. 
economic drop in, 306, 331. 
natural frequency, 303, 336. 
power factor of, 318. 
regulation, 316, 334. 
span lengths, 329, 338. 
structures, 326. 

forces acting on, 341. 
of power, 300. 
Triangle of E.M.F.'s, 38, 64. 
Two-phase power measurement, 107. 
systems, 101. 

Values of inductances, 29. 
Vapor converter, mercury, 296. 
V-curves of synchronous motor, 
»$8, 



INDEX. 



353 



Vector, 5. 

addition and subtraction, 17. 
diagram of transformer, 154, 175, 
178. 
Voltage and current relations in con- 
densive circuit, 63. 
in polyphase systems, 101, 104, 
106. 
armature impedance, 121. 
average value of, 8. 
curves of actual, 6. 
drop in alternators, 117. 

transmission line, 306, 331, 
effective value of, 7. 
for power transmission, 305. 
generated in armature, 96. 

Wagner single-phase motor, 279. 

transformer, 188. 
Water-cooled transformer, 194, 



Wattless component of current, 16, 

159, 230. 
Wattmeter, induction, 246. 
Wave-shape, 3, 

causes of distortion of, 5. 

determination of form factor of, 
10. 

of current in converter coils, 290. 
Weights, relative, of line wire, 305. 
Westinghouse alternator, 126. 

compensated series motor, 271. 

induction motor, 204. 

motor starter, 209. 

transformer, 191. 
Wind pressure on lines, 323. 

Y-connection, 100, 184. 
of compensators, 186. 

Zig-zag leakage reactance, 220. 



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ALLSOP, F. C. Practical Electric Light Fitting. 289 pp. Illustrated. 

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ANDERSON, GEO. L., A. M. (Captain of U. S. Artillery.) Hand-book 

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CROCKER, F. B. Electric Lighting. Vol. I. The Generating Plant. 

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OCT S . 



